# Concept Of Physics Electromagnetic Induction

#### H C Verma

1.   1.Calculate the dimensions of (a) $\int_{}^{}$ $\vec{E}\vec{-dl}$. (b) vBl and (c)$\frac{d\phi_B}{dt}$. $\hspace{0.4cm}$The symbols have their usual meanings

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$\int_{}^{} E.dl =MLT^{-3}l^{-1}\times L=ML^{2}l^{-1}T^{-3}$ $\\$ (b) vBl$=LT^{-1}\times Ml^{-1}T^{-2}\times L=ML^2l^{-1}T^{3}$ $\\$ (c) d$\phi_s/dt=Ml^{-1}T^{-2}\times L^2=ML^2l^{-1}T^{-2}$ $\\$

2.   2.The flux of magnetic field through a closed conducting loop changes with time according to the equation, $\phi$ = $at^2$ + bt + c. (a) Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s

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3.   The flux of magnetic field through a closed conducting loop changes with time according to the equation, $\phi= at^2 + bt + c$. (a) Write the SI units of a, b and c. (b) If the magnitudes of a, b and c are 0.20, 0.40 and 0.60 respectively, find the induced emf at t = 2 s

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$\phi =at^2+bt+c$ $\\$ (a)$\hspace{0.15cm}$ a$=[\frac{\phi}{t^2}]=[\frac{\phi/t}{t}]=\frac{Volt}{Sec}$ $\\$ $b=[\frac{\phi}{t}]=Volt$ $\\$ $c=[\phi]=Weber$ $\\$ (b) $E=\frac{d\phi}{dt} \hspace{0.3cm} [a=0.2,b=0.4,c=0.6,t=2s]$ $\\$ =2at+b $\\$ =2$\times 0.2\times 2+0.4=1.2 volt$ $\\$

4.   3. (a) The magnetic field in a region varies as shown in figure (38-E1). Calculate the average induced emf in a conducting loop of area 2.0 x 10$^{-3}$ m$^2$ placed perpendicular to the field in each of the 10 ms intervals shown, (b) In which intervals is the emf not constant? Neglect the behaviour near the ends of 10 ms intervals.

$\phi_2=B.A=0.01\times2\times10^{-3}=2\times 10^{-5} m$ $\\$ $\phi_1=0$ $\\$ $e=-\frac{d\phi}{dt}=\frac{-2\times10^{-5}}{10\times10^{-3}}=-2mV$ $\\$ $\phi_3=B.A=0.03\times2\times 10^{-3}=6\times 10^{-5}$ $\\$ $d\phi = -4\times 10^{-5}$ $\\$ $e=-\frac{d\phi}{dt}=-4mV$ $\\$ $\phi_4=B.A=0.01\times2\times10^{-3}=2\times 10{^-5}$ $\\$ $d\phi=-4\times10^{-5}$ $\\$ e=$-\frac{d\phi}{dt}=4mV$ $\\$ $\phi_5=B.A=0$ $\\$ $d\phi =-2\times 10^{-5}$ $\\$ $e=-\frac{d\phi}{dt}=2mV$ $\\$ (b) emf is not constant in case of $\rightarrow$ 10-20ms and 20-30 ms as -4mV and 4mV $\\$

5.   4. A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time.

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A=$1mm^2$ ;i=10A, d=20cm;dt=0.1s $\\$ e=$\frac{d\phi}{dt}=\frac{BA}{dt}=\frac{\mu_{\circ}i}{2\pi d}\times\frac{A}{dt}$ $\\$ $=\frac{4\pi\times10^{-7}\times10}{2\pi\times2\times10^{-1}}\times\frac{10^{-6}}{1\times 10^{-1}}=1\times10^{-10}V$ $\\$

6.   5.A conducting circular loop of area 1 mm ~ is placed co-planarly with a long, straight wire at a distance of 20 cm from it. The straight wire carries an electric current which changes from 10 A to zero in 0.1 s. Find the average emf induced in the loop in 0.1 s.

A=$1mm^2$ ;i=10A, d=20cm;dt=0.1s $\\$ e=$\frac{d\phi}{dt}=\frac{BA}{dt}=\frac{\mu_{\circ}i}{2\pi d}\times\frac{A}{dt}$ $\\$ $=\frac{4\pi\times10^{-7}\times10}{2\pi\times2\times10^{-1}}\times\frac{10^{-6}}{1\times 10^{-1}}=1\times10^{-10}V$ $\\$

7.   6. A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a l1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during (a) its removal, (b) its restoration and (c) its motion.

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(a) During removal $\\$ $\phi_1=B.A=1\times50\times 0.5\times 0.5-25\times0.5=12.5Tesla-m^2$ $\\$ $\phi_2=0 ,\tau=0.25$ $\\$ $e=-\frac{d\phi}{dt}=\frac{\phi_2-\phi_1}{dt}=\frac{12.5}{0.25}=\frac{125\times 10^{-1}}{25\times 10^{-2}}=50V$ $\\$ (b) During its restoration $\\$ $\phi_1=0 ; \phi_2=12.5Tesla-m^2 ; t=0.25s$ $\\$ $E=\frac{12.5-0}{0.25}=50V$ $\\$ (c) During the motion $\\$ $\phi_1=0,\phi_2=0$ $\\$ E=$\frac{d\phi}{dt}=0$ $\\$

8.   7. Suppose the resistance of the coil in the preivous problem is 25$\Omega$. Assume that the coil moves with uniform velocity during its removal and restoration. Find the thermal energy developed in the coil during (a) its removal, (b) its restoration and (c) its motion.

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R= 25$\Omega$ $\\$ (a) e=50V, T=0.25s $\\$ i=e/R=2A,H=$i^{2}RT$ $\\$ =$4\times 25\times0.25=25J$ $\\$ (b)e=50 V, T=0.25s. $\\$ i=e/R=2A,$H=i^2RT=25J$ $\\$ (c) Since energy is a scalar quantity $\\$ Net thermal energy developed=25J+25J=50J $\\$

9.   8. A conducting loop of area 5.0 $cm^2$ is placed in a magnetic field which varies sinusoidally with time as B = $B_0$ sin $\omega$t where $B_0 = 0.20$ T and $\omega$ = 300 $s^{-1}$ .The normal to the coil makes an angle of 60° with the field. Find (a) the maximum emf induced in the coil, (b) the emf induced at r = (n/900)s and (c) the emf induced at t = ($\pi$/600) s.

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$A=5cm^2=5\times10^{-4}m^2$ $\\$ $B=B_0sin\omega t=0.2sin(300)t$ $\\$ $\theta=60^{\circ}$ $\\$ a) Max emf induced in the coil , $\\$ E=$-\frac{d\phi}{dt}=\frac{d}{dt}(BA cos\theta)$ $\\$ =$\frac{d}{dt}(B_0sin\omega t\times5\times10^{-4}\times\frac{1}{2})$ $\\$ =$B_0 \times \frac{5}{2}\times10^{-4}\frac{d}{dt}(sin\omega t)=\frac{B_05}{2}\times 10^{-4}cos\omega t . \omega$ $\\$ $E_max=15\times10^{-3}=0.015V$ $\\$ b)Induced emf at t=$(\pi/900)s$ $\\$ E=$15\times10^{-3}\times cos(300\times \pi/600)$ $\\$ $15\times 10^{-3} \times 1/2.$ $\\$ =$0.015/2=0.0075=7.5 \times 10^{-3} V$ $\\$ c) Induced emf at $t=\pi/600$ s E=$1.5 \times 10^{-3}\times cos(300\times \pi/600)$ $\\$ =$15\times 10^{-3} \times 0=0 V$ $\\$

10.   9.Figure (38-E2) shows a conducting square loop placed parallel to the pole-faces of a ring magnet. The pole-faces have an area of 1 cm 2 each and the field between the poles is 0.10 T. The wires making the loop are all outside the magnetic field. If the magnet is removed in1.0 s, what is the average emf induced in the loop ?

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$\vec{B}=0.10T$ $\\$ A=$1cm^2=10^{-4}$ $\\$ T= 1s $\\$ $\phi = B.A=10^{-1}\times10^{-4}=10^{-5}$ $\\$ e=$\frac{d\phi}{dt}=\frac{10^{-5}}{1}=10^{-5}=10\mu V$ $\\$

11.   10. A conducting square loop having edges of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV, find the magnitude of the magnetic field.

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E=$20mV=20\times10^{-3}V$ $\\$ A=$(2\times10^{-2})^2=4\times10^{-4}$ $\\$ Dt=0.2s,$\theta=180^{\circ}$ $\\$ $\phi_1=BA,\phi_2=-BA$ $\\$ $d\phi$=2BA $\\$ E=$\frac{d\phi}{dt}=\frac{2BA}{dt}$ $\\$ $\Rightarrow 20\times 10^{-3}=\frac{2\times B\times 2\times 10^{-4}}{2\times 10^{-1}}$ $\\$ $\Rightarrow 20\times 10^{-3}=4\times B\times 10^{-3}$ $\\$ $\Rightarrow B=\frac{20\times 10^{-3}}{42\times 10^{-3}}=5T$ $\\$

12.   11. A conducting loop of face-area A and resistance R is placed perpendicular to a magnetic field B. The loop is withdrawn completely from the field. Find the charge which flows through any cross-section of the wire in the process. Note that it is independent of the shape of the loop as well as the way it is withdrawn.

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Area =A, Resistance=R, B=Magnetic Field $\\$ $\phi=BA=BAcos0^{\circ}=BA$ $\\$ e=$\frac{d\phi}{dt}=\frac{BA}{1}; i=\frac{e}{R}=\frac{BA}{R}$ $\\$ $\phi=iT=BA/R$ $\\$

13.   12. A long solenoid of radius 2 cm has 100 turn^cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Q is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.

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$r=2cm =2\times10^{-2} m$ $\\$ n=100 turns/cm=10000 turns/m $\\$ i=5A $\\$ B=$\mu_{\circ}ni$ $\\$ =$4\pi \times 10^{-7}\times 10000\times5=20\pi\times10^{-3}=62.8\times10^{-3} T$ $\\$ $n_2=100 turns$ $\\$ R=20$\Omega$ $\\$ r=1cm=$10^{-2} m$ $\\$ Flux linking per turn of the second coil= $B\pi r^2=B\pi \times 10^{-4}$ $\\$ $\phi_1=Total flux linking=Bn_2\pi r^2=100\times \pi \times 10^{-4} \times 20\pi \times 10^{-3}$ $\\$ When current is reversed $\\$ $\phi_2=-\phi_1$ $\\$ $d\phi=\phi_2-\phi_1=2\times 100\times \pi\times 10^{-4} \times 20\pi \times 10^{-3}$ $\\$ $E=-\frac{d\phi}{dt}=\frac{4\pi^{2}\times 10^{-4} }{dt}$ $\\$ $I=\frac{E}{R}=\frac{4\pi^2\times10^{-4}}{dt\times 20}$ $\\$ q=Idt=$\frac{4\pi^2\times10^{-4}}{dt\times 20} \times dt=2\times 10^{-4}C$ $\\$

14.   13. Figure (38-E3) shows a metallic square frame of edge a in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and displace the corners at a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.

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Speed= u $\\$ Magnetic field =B $\\$ Side=a $\\$ a) The perpendicular component i.e $a sin\theta$ is to be taken which is $\perp r$ to velocity. $\\$ So, I=a $sin\theta30^{\circ}=a/2$ $\\$ Net 'a' charge =4\times a/2=2a $\\$ So, induced emf=Bvl=2auB $\\$ b) Current=$\frac{E}{R}=\frac{2auB}{R}$ $\\$

15.   14. The north pole of a magnet is brought clown along the axis of a horizontal circular coil (figure 38-E4). As a result, the flux through the coil changes from 0.35 weber to 0.85 weber in an interval of half a second. Find the average emf induced during this period. Is the induced current clockwise or anticlockwise as you look into the coil from the side of the magnet?

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$\phi_1=0.35 weber, \phi_2=0.85 weber$ $\\$ $D\phi=\phi_2-\phi_1=(0.85-0.35)weber=0.5 weber$ $\\$ dt=0.5 sec $\\$ E=$\frac{d\phi}{dt'}-\frac{0.5}{0.5}=1V$ $\\$ The induced current is therefore anticlockwise $\\$

16.   15. A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.

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$i=v(B\times I)$ $\\$ =vBIcos$\theta$ $\\$ $\theta$ is the angle between normal to plane and $\vec{B}=90^{\circ}$ $\\$ =$vBIcos90^{\circ} = 0$ $\\$

17.   16. Figure (38-E5) shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t =0. Find the emf induced in the loop at (a) t = 2 s, (b) t = 10 s, (c) t = 22 s and (d) t = 30 s.

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$u=1cm/',B=0.6T$ $\\$ at t=10 sec $\\$ distance moved = 10\times 1 = 10cm $\\$ The flux linked does not change with time $\\$ $\therefore E=0$ $\\$ c) At t =22 sec $\\$ distance =$22\times 1 = 22cm$ $\\$ The loop is moving out of the field and 2cm outside $\\$ $E=\frac{d\phi}{dt}=B\times\frac{dA}{dt}$ $\\$ =$\frac{0.6\times\times(2\times5\times 10^{-4})}{2}=3\times 10^{-4} V$ $\\$ d)At t=30sec $\\$ The loop is total outside and flux linked is 0 $\\$ $\therefore E=0$ $\\$

18.   17. Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the resistance of the loop is 4'5 m£2.

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As the heat produced is a scalar prop $\\$ So, net heat produced =$H_a+H_b+H+c+H_d$ $\\$ R=4.5m$\Omega=4.5 \times 10^{-3} \Omega$ $\\$ a)$e=3\times 10^{-4} V$ $\\$ $i=\frac{e}{R}=\frac{3\times 10^{-4}}{4.5\times 10^{-3}}=6.7\times 10^{-2}Amp$ $\\$ $H_a=(6.7\times 10^{-2})^2\times 4.5\times 10^{-3} \times 5$ $\\$ $H_b=H_d=0$ [Since emf is induced for 5 sec] $\\$ $H_c=(6.7\times 10^{-2})^2\times 4.5 \times 10^{-3} \times 5$ $\\$ So total heat = $H_a+H_c=2\times (6.7\times 10^{-2})^2\times 4.5\times 10^{-3}\times 5 =2\times 10^{-4}J.$ $\\$

19.   18. A uniform magnetic field B exists in a cylindrical region of radius 10 cm as shown in figure (38-E6). A uniform wire of length 80 cm and resistance 4.0Q is bent into a square frame and is placed with one side along a diameter of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s, find the current induced in the frame.

r=10cm, r=40cm $\\$ $\frac{dB}{dt}=0.010T/',\frac{d\phi}{dt}=\frac{dB}{dt}A$ $\\$ $E=\frac{d\phi}{dt}=\frac{dB}{dt} \times A=0.01(\frac{\pi \times r^2}{2})$ $\\$ $=\frac{0.01\times 3.14\times 0.01}{2}=\frac{3.14}{2}\times 10^{-4}=1.57\times 10^{-4}$ $\\$ $i=\frac{E}{R}=\frac{1.57\times10^{-4}}{4}=0.39\times 10^{-4}=3.9\times 10^{-5} A$ $\\$
20.   19. The magnetic field in the cylindrical region shown in figure (38-E7) increases at a constant rate of 20.0 mT/s. Each side of the square loop abed and defa has a length of 1.00 cm and a resistance of 4.00 Q. Find the current (magnitude and sense) in the wire ad if (a) the switch S1 is closed but S2 is open, (b)S1 is open but S2 is closed, (c) both S1 and S2 are open and (d) both <$, and S., are closed. Answer 20 None$S_1 $closed,$S_2 open\\$net$ R=4\times 4=16 \Omega\\e=\frac{d\phi}{dt}=A\frac{dB}{dt}=10^{-4}\times 2 \times 10^{-2}=2\times 10^{-6} V \\$i through ad =$ \frac{e}{R}=\frac{2\times 10^{-6}}{16}=1.25\times 10^{-7}A $along ad$\\$b) R=$16\Omega\\$e=A$\times\frac{dB}{dt}=2\times 00^{-5} V \\i=\frac{2\times 10^{-6}}{16}=1.25\times 10^{-7} $A along da$\\$c) Since both$S_1$and$S_2$are open, no current passes i.e i=0$\\$d)Since both$S_1$and$S_2$are closed, the circuit forms a balanced wheat stone bridge and no current will flow along ad i.e i=0.$\\$21. 20. Figure (38-E8) shows a circular coil of N turns and radius a, connected to a battery of emf$\epsilon$through a rheostat. The rheostat has a total length L and resistance R. The resistance of the coil is r. A small circular loop of radius a' and resistance r' is placed coaxially with the coil. The centre of the loop is at a distance x from the centre of coil, the In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards right at a constant speed v. Find the emf induced in the small circular loop at, the instant (a) the contact begins to slide and (b) it has slid through half the length of the rheostat. Answer 21 None Magnetic field due to the coil(1) at the center of (2) is B=$\frac{\mu_{\circ}Nia^2}{2(a^2+x^2)^{3/2}} \\$Flux linked with the second,$\\$=$B.A_{(2)}=\frac{\mu_0Nia^2}{2(a^2+x^2)^{3/2}} \pi ra'^2 \\$E.m.f induced$\frac{d\phi}{dt}=\frac{\mu_{\circ}Na^2a'^2\pi}{2(a^2+x^2)^{3/2}}\frac{di}{dt}\\\frac{\mu_{\circ}N\pi a^2a'^2}{2(a^2+x^2)^{3/2}}\frac{d}{dt}\frac{E}{((R/L)x+r^)} \\=\frac{\mu_{\circ}N\pi a^2a'^2}{2(a^2+x^2)^{3/2}}E-\frac{-1.R/L.v}{(R/L)x+r^)2} \\$b) =$\frac{\mu_{\circ}N\pi a^2a'^2}{2(a^2+x^2)^{3/2}}-\frac{ERV}{L(R/2+r2)^2} \hspace{0.2cm}$(for x=L/2,R/Lx=R/2)$\\$a) For x=L$\\$E=$\frac{\mu_{\circ}N\pi a^2a'^2RvE}{2(a^2+x^2)^{3/2}(R+r)^2}\\$22. 21. A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field B = 0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60.0°. The operation takes 0.100s. (a) Find the average emf induced in the coil, (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00$\Omega$, calculate the net charge crossing a cross-section of the wire of the coil. Answer 22 None N=50$\vec{B}$=0.200 T; r=200c,=0.02cm$\\\theta=60^{\circ}$, t=0.100s$\\$a) e=$\frac{Nd_{phi}}{dt}=\frac{n\times B.A}{T}=\frac{NBAcos60^{\circ}}{T} \\=\frac{50\times2\times10^{-1}\times \pi \times(0.02)^2}{0.1}=5\times4\times10^{-3} \times \pi \\$=$2 \pi \times 10^{-2}V=6.28\times 10-^{-2}V \\$b) i=$\frac{e}{R}=\frac{6.28\times 10^{-2}}{4}=1.57\times 10^{-3} C \\$23. A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 x 10 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm{ ^{2 }and its resistance is 4.0 Q. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a). Answer 23 None r=10cm=0.1m$\\$R=40$\Omega,N=1000 \\\theta=180^{\circ},B_H=3\times 10^{-5} T \\\phi=N(B.A)=NBA Cos 180^{\circ} or =-NBA \\$=$1000\times3\times10^{-5}\times \pi \times 1\times10^{-2}=3\pi \times 10^{-4} where \\d\phi=2NBA=6\pi \times 10^{-4} weber \\e=\frac{d\phi}{dt}=\frac{6\pi\times10^{-4}V}{dt} \\$i=$\frac{6\pi\times10^{-4}}{40dt}=\frac{4.71\times 10^{-5}}{dt} \\$Q=$\frac{4.71\times10^{-5}\times dt}{dt}4.71\times 10^{-5} C. \\$24. A coil of radius 10 cm and resistance 40 Q has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer if the horizontal component of the Earth's magnetic field is$B_H=3.0\times10^{-5} T. \\$Answer 24 None r=10cm=0.1m$\\$R=40$\Omega,N=1000 \\\theta=180^{\circ},B_H=3\times 10^{-5} T \\\phi=N(B.A)=NBA Cos 180^{\circ} or =-NBA \\$=$1000\times3\times10^{-5}\times \pi \times 1\times10^{-2}=3\pi \times 10^{-4} where \\d\phi=2NBA=6\pi \times 10^{-4} weber \\e=\frac{d\phi}{dt}=\frac{6\pi\times10^{-4}V}{dt} \\$i=$\frac{6\pi\times10^{-4}}{40dt}=\frac{4.71\times 10^{-5}}{dt} \\$Q=$\frac{4.71\times10^{-5}\times dt}{dt}4.71\times 10^{-5} C. \\$25. 24. A circular coil of one turn of radius 5.0 cm is rotated about a diameter about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.010T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the average emf induced in the coil over a long period and (c) the average of the squares of emf induced over a long period Answer 25 None$emf=\frac{d\phi}{dt}=\frac{dB.Acos\theta}{dt} \\$=BAsin$\theta\omega=-BA\omega sin\theta\\$(dq/dt=the rate of change of angle between arc vector and B=$\omega$)$\\$a) emf maximum=$BA_{\omega}=0.010\times25\times10^{-4}\times 80 \times \frac{2\pi\times\pi}{6} \\$=$0.66\times10^{-3}=6.66\times 10^{-4} \\$b) Since the induced emf changes its direction every time,$\\$the average emf=0$\\$26. 25.Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Q. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute. Answer 26 None H=$\int_{0}^{t}i^{2}Rdt=\int_{0}^{t}\frac{B^{2}A^{2}\omega^{2}}{R^2} sin\omega t R dt \\$=$\frac{B^{2}A^{2}\omega^{2}}{2R^{2}}\int_{0}^{t}(1-cos2\omega t)dt \\$=$\frac{B^{2}A^{2}\omega^{2}}{2R}(t-\frac{sin2\omega t}{2\omega})_{0}^{1minute} \\=\frac{B^2A^2\omega^2}{2R}(60-\frac{sin2\times8-x2\pi/60\times 60}{2\times 80\times 2\pi/60}) \\\frac{60}{200} \times \pi^2r^4\times B^2\times (80^4\times\frac{2\pi}{60})^2 \\$=$\frac{60}{200}\times 10\times \frac{64}{9} \times 10 \times 625 \times 10^{-8} \times 10^{-4}=\frac{625\times6\times 64}{9\times 2}\times 10^{-11}=1.33\times 10^{-7} J \\$27. 26.Figure (38-E9) shows a circular wheel of radius 10.0 cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field B of magnitude$2.00 \times 10^{-4}$T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period. Answer 27 None$\phi=BA,\phi_2=0\\$=$\frac{2\times 10^{-4} \times \pi(0.1)^2}{2}=\pi\times 10^{-5} \\$E=$\frac{d\phi}{dt}=\frac{\pi\times 10^{-6}}{2}=1.57\times 10^{-6} V. \\$28. 27.A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 city's perpendicular to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion, (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force ? How is this electric field created ? (c)Find the motional emfs between the ends of the rod. Answer 28 None$I=20cm=0.2m \\$v=10cm/s=0.1m/s$\\$B=0.10T$\\$a)$ F=qvB=1.6\times 10^{-19}\times 1 \times 10^{-1} \times 1 \times 10^{-1}=1.6\times 10^{-21} N\\$b)qE=qvB$\\\Rightarrow E=1\times 10^{-1}\times 1 \times 10^{-1}= 1\times 10^{-2} V/m \\$This is created due to the induced emf$\\$c) Motional emf= Bvl$\\=0.1\times0.1\times 0.2=2\times 10^{-3} V \\$29. 28.A metallic metre stick moves with a velocity of 2 m/s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.2 T. Find the emf induced between the ends of the stick. Answer 29 None l=1m,B=0.2T,v=2m/s,e=Blv$\\$=$0.2\times 1 \times 2=0.4 V\\$30. 29.A 10 m wide spacecraft moves through the interstellar space at a speed 3 x 10$^{7}$m/s. A magnetic field B= 3$\times$10$^{10}$T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.$l=10m, V=3$\times 10^7$ m/s, B=$3\times 10^{-10}T$ $\\$ Motional emf=Bvl $$\\ =3\times 10^{-10} \times 3 \times 10^{7}\times 10=9\times 10^{-3}=0.09V  \\ 31. 30. The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180 km/h? The vertical component of earth's magnetic field is 0.2 \times 10 ^{-4}\hspace{0.1cm}T and the rails are separated by 1 m. v=180 km/h=50m/s  \\ B=0.2 \times 10^{-4} T, L= 1m  \\ E=Bvl=0.2 I 10^{-4} \times 50=10^{-3} V  \\ \therefore the voltmeter will record 1 mv. \\ 32. 31.A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in b ©Bfigure (38-E10). A uniform magnetic field B exists in the" perpendicular direction. Find the emf induced (a) in the loop abc, (b) in the segment bc, (c) in the segment ac and (d) in the segment ab. a) Zero as the components of ab are exactly opposite to that of bc. So they cancel each othher. Because velocity should be perpendicular to the length. b) e=Bv\times l  \\ =Bv(bc)+ve at C \\ c) e=0 as the velocity is not perpendicular to the length \\ d)e=Bv(bc) positive at 'a' \\ i.e the component of 'ab' along the perpendicular direction \\ 33. 32. A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joining free ends, (b) the velocity is parallel to this diameter. a) The component of length moving perpendicular to V is 2R \\ \therefore E=B\hspace{0.15cm}v\hspace{0.15cm} 2R  \\ b) Component of length perpendicular to velocity 0 \\ \therefore E=0  \\ 34. 33. A wire of length 10 cm translates in a direction making an angle of 60° with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in the space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm/s l=10cm=0.1m;  \\ \theta=60^{\circ}; B=1T  \\ V=20 cm/s=0.2m/s \\ E=Bvlsin60^{\circ} \\ [As we have to take the component of length vector which is \perp r to the velocity vector] \\ =1\times0.2\times0.1\times\sqrt{3}/2  \\ =1.732\times10^{-2}=17.32 \times 10^{-3}V  \\ 35. 34. A circular copper-ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring, (a) Between: which pair of points is the emf maximum ? What is the value of this maximum emf? (b) Between which pair of points is the emf minimum ? What is the value of this minimum emf? a) The e.m.f is highest between diameter \perpr to the velocity vector. Because here length \perpr to the velocity is highest. \\ E_{max}=VB2R  \\ b) The length perpendicular to the velocity is lowest as the diameter is parallel to the velocity E_{min}=0.  \\ 36. 35. Figure (38-E11) shows a wire sliding on two parallel, conducting rails placed at a separation L A magnetic field B exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity v ? F_{magnetic}=ilB  \\ This force produces an acceleration of the wire. \\ But since the velocity is given to be constant. \\ Hence net force acting on the wire must be zero \\ 37. 36 . Figure (38-E 12) shows a long U-shaped wire of width I placed in a perpendicular magnetic field B. A wire of length I is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit. E= Bvl \\ Resistance= r \times total length \\ =r\times 2(l+vt)=24(l+vt)  \\ i=\frac{Bvl}{2r(l+vl)}  \\ 38. 37.Consider the situation of the previous problem, (a) Calculate the force needed to keep the sliding wire moving with a constant velocity v. (b) If the force needed just after t = 0 is F_0, find the time at which the force needed will be F_0 /2. e=Bvl \\ i=\frac{e}{R}=\frac{Bvl}{2r(l+vl)}  \\ a) F=ilB=\frac{Bvl}{2r(l+vl)}\times lB=\frac{B^{2}l^2v}{2r(l+vt)}  \\ b) Just after t=0 \\ F_0=ilB=lB(\frac{lBv}{2rl})=\frac{lB^2v}{2r}  \\ \frac{F_0}{2}=\frac{lB^2v}{4r}=\frac{l^2B^2v}{2r(l+vl)}  \\ \Rightarrow 2l=l+vt  \\ \Rightarrow T=l/v  \\ 39. 38.Consider the situation shown in figure (38-E13). The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed u0. Find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instant, (c) the velocity v as a function of x and (d) the maximum distance the wire will move. a) When the speed is V \\ Emf=Blv \\ Resistance =r+r \\ Current =\frac{Blv}{r+R}  \\ b) Force acting on the wire=ilB \\ =\frac{BlvlB}{R+r}=\frac{B^2l^2v}{R+r}  \\ Acceleration on the wire=\frac{B^2l^2v}{m(R+r)}  \\ c) v=v_{\circ}+at=v_0-\frac{B^2l^2v}{m(R+r)}  [Force is opposite to velocity]\\ =v_0-\frac{B^2l^2x}{m(R+r)}  \\ d)\hspace{0.15cm}a=v\frac{dv}{dx}=\frac{B^2l^2v}{m(R+r)}  \\ \Rightarrow dx=\frac{dvm(R+r)}{B^2l^2}  \\ \Rightarrow x=\frac{m(R+r)v_0}{B^2l^2}  \\ 40. 39. A rectangular frame of wire abcd has dimensions 32 cm \times 8 .0 cm and a total resistance of 2.0 \Omega. It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2 x 10 ^{-5} N (figure 38-E14). It is found that the frame moves with constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points a and b and (d) the potential difference between the points c and d. R=2.0\Omega,B=0.020T, I=32cm=0.32m \\ B=8cm=0.08 m \\ a)F=ilB=3.2\times 10^{-5} N  \\ =\frac{B^2l^2v}{R}=3.2\times 10^{5}  \\ \Rightarrow \frac{(0.020)^2 \times (0.08)^2\times v}{2}=3.2\times 10^{-5}  \\ \Rightarrow v=\frac{3.2\times 10^{-5}\times 2}{6.4\times 10^{-3}\times 4\times 10^{-4}}=25m/s  \\ b)Emf E= vBl=25\times 0.02\times 0.08=4\times 10^{-2} v.  \\ c) Resistance per unit length=\frac{2}{0.8} \\ Resistance of part ad/cb=\frac{2\times 0.72}{0.8}=1.8\Omega \\ V_{ab}=iR=\frac{Blv}{2} \times 1.8=\frac{0.02\times0.08\times25\times 0.2}{2}=0.036V=3.6\times10^{-2} V  \\ d) Resistance of cd=\frac{2\times0.08}{0.8}=0.2\Omega \\ V=iR=\frac{0.02\times0.08\times25\times 0.2}{2}=4\times 10^{-3}V  \\ 41. 40.Figure (38-E15) shows a metallic wire of resistance 0.20 \Omega sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 \mu A passes through the wire when it is slid at a rate of 20 cm/s. If the horizontal component of the earth's magnetic field is 3.0 \times 10^{- 5} \hspace{0.1cm}T, calculate the dip at the place. I=\frac{Blv}{R}=\frac{B\times l\times cos\theta\times vcos\theta}{R}  \\ =\frac{Blv}{R}cos^2\theta \\ F=ilB=\frac{Blv cos^2\theta\times lB}{R} \\ Now F=mgson\theta [Force due to gravity which pulls downwards] \\ Now,\frac{B^2l^2vcos^2\theta\times \theta}{R}=mgsin\theta  \\ \Rightarrow B=\sqrt{\frac{mg Rsin\theta}{vl^2cos^2\theta}} 42. 41. A wire ab of length /, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure (38-E16). The plane of the rails makes an angle \theta with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that \sqrt{\frac{mg Rsin\theta}{vl^2cos^2\theta}} I=\frac{Blv}{R}=\frac{B\times l\times cos\theta\times vcos\theta}{R}  \\ =\frac{Blv}{R}cos^2\theta \\ F=ilB=\frac{Blv cos^2\theta\times lB}{R} \\ Now F=mgson\theta [Force due to gravity which pulls downwards] \\ Now,\frac{B^2l^2vcos^2\theta\times \theta}{R}=mgsin\theta  \\ \Rightarrow B=\sqrt{\frac{mg Rsin\theta}{vl^2cos^2\theta}} 43. Consider the situation shown in figure (38-E17). The wires P_1Q_2 and P_2Q_2 are made to slide on the rails with the same speed 5 cm/s. Find the electric current in the 19 \Omega resistor if (a) both the wires move towards right and (b) if P_1Q_1 moves towards left but P_2Q_2 moves towards right. a) The wire constitutes 2 parallel emf. \\ \therefore \hspace{0.2cm}Net \hspace{0.2cm}emf=B l v=1\times4\times10^{-2}\times 5\times 10^{-2}=20\times 10^{-4}  \\ Net resistance=\frac{2\times 2}{2+2}+19=20\Omega \\ Net current=\frac{20\times10^{-4}}{20}=0.1mA  \\ b) When both the wires move towards opposite directions then not emf=0 \\ \therefore Net current=0 44. 43. Suppose the 19 CI resistor of the previous problem is disconnected. Find the current through P_2Q_2 in the two situations (a) and (b) of that problem. a) No current will pass as the circuit is incomplete \\ b) As the circuit is complete \\ VP_2Q_2=Blv \\ =1\times0.04\times 0.05=2\times 10^{-3} V \\ R=2\Omega \\ i=\frac{2\times10^{-3}}{2}=1\times 10^{-3}A=1mA \\ 45. 44.Consider the situation shown in figure (38-E18). The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm/s. Find the current in the 10 \Omega resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail. B=1 T,V=5|10^{-2}m/',R=10\Omega \\ a) When the switch is thrown to the middle rail \\ E=Bvl\\ =1\times5\times10^{-2}\times 2\times 10^{-2}=10^{-3}  \\ Current in the 10\Omega resistor=E/R \\ =\frac{10^{-3}}{10}=10^{-4}=0.1mA \\ b) The switch is thrown to the lower rail \\ E=Bvl \\ =1\times5\times10^{-2}\times 2\times 10^{-2}=20\times 10^{-4}  \\ Current=\frac{20\times 10^{-4}}{10}=2\times 10^{-4}=0.2mA  \\ 46. 45. The current generator Ig, shown in figure (38-E20), sends a constant current i through the circuit. The wire ab has a length I and mass m and can slide on the smooth, horizontal rails connected to Ig . The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time. Initial current passing=i \\ Hence initial emf=ir \\ Emf due to motion of ab=Blv \\ Net emf=ir-Blv\\ Net resistance=2r \\ Hence current passing=\frac{ir-Blv}{2r} \\ Force on the wire=ilB \\ Acceleration=\frac{ilB}{m} \\ Velocity=\frac{ilBt}{m} \\ 47. 46. The current generator Ig, shown in figure (38-E20), sends a constant current i through the circuit. The wire ab has a length I and mass m and can slide on the smooth, horizontal rails connected to Ig. The enitre system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time. Force on the wire=ilB \\ Acceleration=\frac{ilB}{m} \\ Velocity=\frac{ilBt}{m} \\ 48. 47. The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails (figure 38-E21). It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass another wire of double its mass, how long will it take in falling through a distance equal to its length ? Given Blv=mg ...(1) \\ When wire is replaced we have \\ 2mg-Blv=2ma\hspace{0.25cm} [Where a\rightarrowacceleration] \\ \Rightarrow a=\frac{2mg-Blv}{2m} \\ Now, s=ut+\frac{1}{2}at^2 \\ l=\frac{1}{2}\times\frac{2mg-Blv}{2m}\times t^2 [\therefore s=l] \\ \Rightarrow t=\sqrt{\frac{4ml}{2mg-Blv}}=\sqrt{\frac{4ml}{2mg-mg}}=\sqrt{2lIg}.[from (1) ] \\ 49. The rectangular wire-frame, shown in figure (38-E22), has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t = 0. (a) Find the acceleration of the frame when its speed has increased to v. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity v_0 \\. (c) Show that the velocity at time is given by v=v_0(1-e^{ft/mv_0}). emf developed=Bdv( when it attains speed v) \\ Current=\frac{Bdv}{R}  \\ Force=\frac{Bd^2v^2}{R}  \\ This force opposes the given force \\ Net F=F-\frac{Bd^2v^2 }{R}=RF-\frac{Bd^2v^2}{R} \\ Net acceleration=\frac{RF-B^2d^2v}{mR} \\ b) Velocity becomes constant when acceleration is 0.\\ \frac{F}{m}-\frac{B^2d^2v_0}{mR}=0$$\\\Rightarrow \frac{F}{m}=\frac{B^2d^2v_0}{mR} \\\Rightarrow V_0=\frac{FR}{B^2d^2} \\$c) Velocity at line t$\\$a=-$\frac{dv}{dt}\\\Rightarrow \int_{0}^{v}\frac{dv}{RF-I^2B^2v}=\int_{0}^{t}\frac{dt}{mR}\\\Rightarrow [I_nRF-I^2B^2V]\frac{1}{-I^2B^2}]_{0}^{v}\hspace{0.2cm} [\frac{t}{Rm}]_{0}^{t} \\\Rightarrow [I_n(RF-I^2B^2v)]_{0}^{v}=\frac{-tl^2B^2}{Rm} \\\Rightarrow I_n(RF-I^2B^2v)-I_n(RF)=\frac{-t^2B^2t}{Rm} \\\Rightarrow 1-\frac{I^2B^2v}{RF}=e^{\frac{-i^2B^2t}{Rm}} \\\Rightarrow \frac{I^2B^2v}{RF}=1-e^{\frac{-i^2B^2t}{Rm}} \\\Rightarrow v=\frac{FR}{I^2B^2}(1-e^{{-i^2B^2v_0t}{Rv_0m}})=v_0(1-e^{-Fv_0m})\\$50. 49. Figure (38-E23) shows a smooth pair of thick metallic rails connected across a battery of emf$\epsilon$having a negligible internal resistance. A wire ab of length I and resistance r can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity v towards right, (a) Find the current in it at this instant. What is the direction of the current ? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire ab will slide with a constant velocity. Find this velocity. Answer 50 None Net emf=E-Bvl$\\$I=$\frac{E-Bvl}{r}$from b to a$\\$F=Ilb$\\$=$(\frac{E-Bvl}{r})lB=\frac{lB}{r}(E-Bvl) $towards right.$\\$After some time when E=Bvl,$\\$Then the wire moves with constant velocity v$\\$Hence v=E I Bl.$\\$51. 50. A conducting wire ab of length I, resistance r and mass m starts sliding at t = 0 down a smooth, vertical, thick pair of connected rails as shown in figure (38-E24). A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails, (a) Write the induced emf in the loop at an instant t when the speed of the wire is v. (b) What would be the magnitude and direction of the induced current in the wire ? (c) Find the downward acceleration of the wire at this instant, (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity$v{_m}$. (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a function of time, (g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached. a) When the speed of wire is V$\\$emf developed=BlV$\\$b) Induced current in the wire=$\frac{BlV}{R}$(from b to a)$\\$c) Down ward acceleration of the wire$\\$=$\frac{mg-F}{m} $due to the current$\\$=mg-ilB/m=g-$\frac{B^2l^2V}{Rm} \\$d) Let the wire start moving with constant velocity. Then acceleration = 0$\\\frac{B^2l^2v}{Rm}m=g\\\Rightarrow V_m=\frac{gRm}{B^2l^2}\\$e)$\frac{dV}{dt}=a \\ \Rightarrow \frac{dV}{dt}=\frac{mg-B^2l^2vR}{m} \\\Rightarrow \frac{dv}{\frac{mg-B^2l^2v/R}{m}}=dt \\\Rightarrow \int_{0}^{v} \frac{mdv}{mg-\frac{B^2l^2v}{R}}=\int_{0}^{t} dt \\\Rightarrow \frac{m}{\frac{-B^2l^2}{R}}(log(mg-\frac{B^2l^2v}{R})_{0}^{v}=t \\\Rightarrow \frac{-mR}{B^2l^2}=log[log(mg-\frac{B^2l^2v}{R})-log(mg)]=t \\\Rightarrow log[\frac{mg-\frac{B^2l^2v}{R}}{mg}]=\frac{-tB^2l^2}{mR} \\\Rightarrow log[1-\frac{B^2l^2v}{Rmg}]=\frac{-tB^2l^2}{mR} \\\Rightarrow 1-\frac{B^2l^2v}{Rmg}=e^{\frac{-tB^2l^2}{mR}} \\\Rightarrow (1-e^{B^2l^2/mR} )=\frac{B^2l^2v}{Rmg} \\\Rightarrow v=\frac{Rmg}{B^2l^2}(1-e^{B^2l^2/mR}) \\\Rightarrow v=v_m(1-e^{-gt/Vm}) [v_m=\frac{Rmg}{B^2l^2}] \\$f)$\frac{ds}{dt}=v\Rightarrow ds=v dt \\\Rightarrow s =vm \int_{0}^{t}(1-e^{-gt/vm}) dt \\$=$V_m(t-\frac{V_m}{g}e^{-gt/vm})=(V_mt+\frac{V_m^{2}}{g}e^{-gt/vm})-\frac{V_m^{2}}{g} \\$=$V_{m}t-\frac{V_m^2}{g}(1-e^{-gt/vm})\\$g)$\frac{d}{dt}mgs=mg\frac{ds}{dt}=mgV_m(1-e^{-gt/vm}) \\\frac{d_H}{dt}=i^2R=R{(\frac{lBV}{R})}^2=\frac{l^2B^2v^2}{R} \\\Rightarrow \frac{l^2B^2}{R}V_m^{2}{(1-e^{-gt/vm})}^2 \\$After steady state i.e T$\rightarrow \infty \\\frac{d}{dt} mgs=mgV_{m} \\\frac{d_H}{dt}=\frac{l^2B^2}{R}V_m^{2}=\frac{l^2B^2}{R}V_{m}\frac{mgR}{l^2B^2}=mgV{m} \\$Hence after steady state$\frac{d_H}{dt}=\frac{d}{dt}mgs \\$52. 51. A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of 100 revolutions per minute. If the length of each spoke-is 30.0 cm and the horizontal component of the earths magnetic field is$2.0 \times 10^{-5} $T, find the emf induced between the axis and the outer end of a spoke. Neglect centripetal force acting on the free electrons of the spoke.$l$=0.3cm,$\vec{B}=2.0 \times 10^{-5} T ,\omega=100 rpm \\v=\frac{100}{60}\times 2\pi=\frac{10}{3}\pi rad/s \\v=\frac{$l$}{2}\ times \omega=\frac{0.3}{2} \times\frac{10}{3}\pi \\$Emf=e=B$l$v$\\$=$2.0 \times 10^{-5} \times 0.3 \times \frac{0.33}{2} \times \frac{10}{3}\pi \\$=$3\pi \times 10^{-6}V=3 \times 3.14 \times 10^{-6}V=9.42 \times 10^{-6} V \\$53. 52. A conducting disc of radius r rotates with a small but constant angular velocity co about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery of the disc. V at a distance r/2$\\$From the centre =$\frac{r\omega}{2} \\$E=B$l$v$\Rightarrow E=B\times r \times \frac{r\omega}{2}=\frac{1}{2} Br^2\omega \\$54. 53. Figure (38-E25) shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor of resistance R is connected between the centre and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the centre ? The radius of the disc is 5'0 cm, angular speed$\omega$= 10 rad/s, B = 0.40 T and R =10$\Omega$.$B=0.40 T, \omega=10 rad/',r=10\Omega \\$r=5cm=0.05m$\\$Considering a rod of length 0.05m affixed at the centre and rotating with the same$\omega \\v=\frac{l}{2} \times \omega =\frac{0.05}{2} \times 10 \\e=Blv=0.40\times \frac{0.05}{2} \times 10 \times 0.05=5 \times 10^{-3} \\I=\frac{e}{R}=\frac{5\times10^{-3}}{10}=0.5mA \\$It leaves from the centre$\\$55. 54. The magnetic field in a region is given by$\vec{B}$=$\vec{k}\frac{B_{\circ}}{L}\gamma$where L is a fixed length. A conducting rod of length L lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity$v = v_{\circ}\vec{i} \\$, find the emf induced between the ends of the rod.$\vec{B}=\frac{B_{0}}{L}y\hat {K} \\$L= Length of rod on y-axis$\\V=V_{\circ}\hat{i} \\$Considering a small length by the rod$\\$dE=B V dy$\\\Rightarrow dE=\frac{B_{\circ}}{L} y\times V_{\circ} \times dy \\\Rightarrow dE=\frac{B_{\circ}V_{\circ}}{L} ydy \\\Rightarrow E=\frac{B_{\circ}V_{\circ}}{L} \int_{0}^{L}ydy \\=\frac{B_{\circ}V_{\circ} }{L}[\frac{y^2}{2}]_{0}^{L}=\frac{B_{\circ}V_{\circ}}{L}\frac{L^2}{2}=\frac{1}{2}B_{\circ}V_{\circ} L \\$56. 55. Figure (38-E26) shows a straight, long wire carrying a current i and a rod of length I coplanar with the wire and perpendicular to it. The rod moves with a constant velocity$v$in a direction parallel to the wire. The distance of the wire from the centre of the rod is x. Find the motional emf induced in the rod. In this case$\vec{B}$varies$\\$Hence considering a small element at centre of rod length dx at a dist x from the wire.$\\\vec{B}=\frac{\mu_{0}i}{2\pi x} \\$So, de=$\frac{\mu_{\circ}i}{2\pi x} \times vxdx \\e=\int_{0}^{e}de=\frac{\mu_{\circ}iv}{2\pi}=\int_{x-t/2}^{x+t/2} \frac{dx}{x}=\frac{\mu_{\circ}iv}{2\pi}[ln(x+ l/2)-ln(x-l/2)] \\$=$\frac{\mu_{\circ}iv}{2}ln[\frac{x+l/2}{x-l/2}]=\frac{\mu_{\circ}iv}{2x}ln(\frac{2x+l}{2x-l}) \\$57. 56.Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance R. (a) What force is needed to keep the rod sliding at a constant speed$v$? (b) In this situation what is the current in the resistance R ? (c) Find the rate of heat developed in the resistor, (d) Find the power delivered by the external agent exerting the force on the rod. Answer 57 None a) emf produced due to the current carrying wire =$\frac{\mu_{\circ}iv}{2\pi}ln(\frac{2x+l}{2x-l}) \\$Let the current produced in the rod =i'=$\frac{\mu_{\circ}iv}{2\pi R}ln(\frac{2x+l}{2x-l}) \\$Force on the wire considering a small portion dx at a distance x$\\dF=i' B l\\\Rightarrow dF=\frac{\mu_{\circ}iv}{2\pi R}ln(\frac{2x+l}{2x-l})\times \frac{\mu_{\circ}i}{2\pi x} \times dx \\\Rightarrow dF=(\frac{\mu_{\circ}i}{2\pi})^2 \frac{v}{R}ln(\frac{2x+l}{2x-l})\frac{dx}{x} \\\Rightarrow F=(\frac{\mu_{\circ}i}{2\pi})^2 \frac{v}{R}ln(\frac{2x+l}{2x-l}) \int_{x-t/2}^{x+t/2}\frac{dx}{x} \\$58. 57.Figure (38-E27) shows a square frame of wire having a total resistance r placed coplanarly with a long, straight wire. The wire carries a current i given by i = i$_{\circ}sin\omega t$. Find (a) the flux of the magnetic field through the square frame, (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to$\frac{20\pi}{\omega}\\$Considering an element dx at a dist x from the wire. We have$\\\phi=B.A\\d\phi=\frac{\mu_{\circ}i\times adx}{2\pi x} \\\phi=\int_{0}^{a}d\phi=\frac{\mu_{\circ}ia}{2\pi}\int_{b}^{a+b}\frac{dx}{x}=\frac{\mu_{\circ}ia}{2\pi}ln{1+a/b} \\$b) e=$\frac{d \phi}{dt}=\frac{d}{dt}\frac{\mu_{\circ}ia}{2\pi}[1+a/b] \\$=$\frac{\mu_{\circ}a}{2\pi}ln[1+a/n]\frac{d}{dt}(i_{\circ}sin\omega t) \\$=$\frac{\mu_{\circ}ai_{\circ}\omega cos \omega t}{2\pi}ln[1+a/b] \\$c)$i=\frac{e}{r}=\frac{\mu_{\circ}ai_{\circ}\omega cos\omega t}{2\pi}ln[1+a/b] \\H=i^{2}rt \\$=${[\frac{\mu_{\circ}ai_{\circ}\omega cos \omega t}{2\pi r}ln(1+a/b)]}^2\times r \times t \\=\frac{\mu_{\circ}^2\times a^2\times i_{\circ}^2 \times \omega^2}{4\pi \times r^2}ln^2[1+a/b]\times r \times \frac{20\pi}{\omega} \\=\frac{5\mu_{\circ}^2a^2i_{\circ}^2\omega}{2\pi r} ln^2[1+a/b] [\therefore t=\frac{20\pi}{\omega} ]\\$59. 58.A rectangular metallic loop of length I and width b is placed coplanarly with a long wire carrying a current i (figure 38-E29). The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance a from the wire. Solve by using Faraday's law for the flux through the loop and also by replacing different segments with equivalent batteries. a) Uising Faraday's Law$\\$Consider a unit length dx at a distance x$\\B=\frac{\mu_{\circ}i}{2\pi x} \\$Area of strip=b dx$\\d\phi=\frac{\mu_{\circ}i}{2\pi x}dx \\\Rightarrow \phi=\int_{a}^{a+I}\frac{\mu_{\circ}i}{2\pi x} bdx \\=\frac{\mu_ {\circ}i}{2\pi}b\int_{a}^{a+I}(\frac{dx}{x})=\frac{\mu_{circ}ib}{2\pi}log(\frac{a+I}{a}) \\Emf=\frac{d\phi}{dt}=\frac{d}{dt}[\frac{\mu_{\circ}ib}{2\pi}log(\frac{a+I}{a})] \\$=$\frac{\mu_{\circ}ib}{2\pi}\frac{a}{a+I}(\frac{va-(a+I)v}{a^2}) $(where da/dt=V)$\\$=$\frac{\mu_{\circ}ib}{2\pi}\frac{a}{a+I}\frac{vI}{a^2}=\frac{\mu_{\circ}ibvl}{2\pi(a+I)a} \\$The velocity of AB and CD creates the emf. Since the emf due to AD and BC are equal and opposite to each other.$\\B_{AB}=\frac{\mu_{\circ}i}{2\pi a} \hspace{0.3cm} \Rightarrow Emf A.B=\frac{\mu_{\circ}i}{2\pi a}bv \\$Length b, velocity v.$\\B_{CD}=\frac{\mu_{\circ}i}{2\pi(a+I)} \\\Rightarrow E.m.f CD=\frac{\mu_{\circ}ibv}{2\pi(a+I)} \\$Length b, velocity v.$\\$Net emf=$\frac{\mu_{\circ}i}{2\pi a}bv-\frac{\mu_{\circ}ibv}{2\pi (a+I)}=\frac{\mu_{\circ}ibvI}{2\pi a(a+I)} \\$60. 59. Figure (38-E29) shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity co. Find the force.$e=Bvl=\frac{B\times a \times \omega \times a}{2} \\i=\frac{Ba^{2}\omega}{2R} \\$F=$ilB=\frac{Ba^2\omega}{2R}\times a \times B=\frac{B^2a^3\omega}{2R} $towards the right of OA.$\\$61. 60.Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made to rotate with a uniform angular speed to as shown in the figure. Find the current in the rod when$\angle$AOC =90°. Answer 61 None 62. 61.Consider a variation of the previous problem (figure 38-E29). Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity co in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle 6 with the vertical. We know$\\F=\frac{B^2a^2\omega}{2R}=ilB \\$Component of mg along F=mgsin$\theta\\$Net force=$\frac{B^2a^3\omega}{2R}-mgsin\theta. \\$63. 62. Figure (38-E30) shows a situation similar to the previous problem. All parameters are the same except that a battery of emf £ and a variable resistance R are connected between O and C. The connecting wires have zero resistance. No external force is applied on the rod (except gravity, forces by the magnetic field and by the pivot). In what way should the resistance R be changed so that the rod may rotate with uniform angular velocity in the clockwise direction? Express your answer in terms of the given quantities and the angle 0 made by the rod OA with the horizontal. emf=$\frac{1}{2}B\omega a^2 \hspace{0.2cm} $[from previous question]$\\$Current=$\frac{e+E}{R}=\frac{1/2\times B\omega a^2 + E}{R}=\frac{B\omega a^2+2E}{2R} \\\Rightarrow mgcos\theta=ilB $[Net force acting on the rod is O]$\\\Rightarrow mgcos\theta =\frac{B\omega a^2+2E}{2R} a\times B \\\Rightarrow R=\frac{(B\omega a^2+2E)aB}{2mgcos\theta} \\$64. 63. A wire of mass m and length I can slide freely on a pair of smooth, vertical rails (figure 38-E31). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance. Let the rod have a velocity v at any instant,$\\$Then at the point,$\\e=Blv \\$Now,$q=c\times potential=ce=CBlv \\$Current I=$\frac{dq}{dt}=\frac{d}{dt}CBIv \\$=$CBI\frac{dv}{dt}=CBIa \hspace{0.25cm} $( where a$\rightarrow$acceleration)$\\$From the figure, force due to the magnetic field and gravity are opposite to each other.$\\$So,$mg-IlB=ma\\\Rightarrow mg-CBla\times lB=ma \hspace{0.3cm} \Rightarrow ma+CB^2l^2a=mg \\\Rightarrow a(m+CB^2l^2)=mg \hspace{0.4cm} \Rightarrow a=\frac{mg}{m+CB^2l2} \\$65. 64. A uniform magnetic field B exists in a cylindrical region, shown dotted in figure (38-E32). The magnetic field increases at a constant rate$\frac{dB}{dt}$. Consider a circle of radius r coaxial with the cylindrical region, (a) Find the magnitude of the electric field E at a point on the circumference of the circle, (b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude found in part (a). a) Work done per unit test change$\\$=$\phi E.dl$(E=electric field)$\\\phi E.dl=e \\\Rightarrow E_{\phi}dl=\frac{d\phi}{dt} \Rightarrow E2\pi r=\frac{dB}{dt}\times A \\\Rightarrow E2\pi r=\pi r^2\frac{dB}{dt} \\\Rightarrow E=\frac{\pi r^2}{2\pi}\frac{dB}{dt}=\frac{r}{2}\frac{dB}{dt} \\$b) When the square is considered,$\\\phi E dl= e \\\Rightarrow E\times 2r\times 4=\frac{dB}{dt}(2r)^2 \\\Rightarrow E=\frac{dB}{dt}\frac{4r^2}{8r} \Rightarrow E=\frac{r}{2}\frac{dB}{dt} \\\therefore$The electric field at the point p has the same value as(a)$\\$66. 65.The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. The solenoid has 2000 turns/m and its radius is 6.0 cm. (a) Consider a circle of radius l.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds, (b) Find the electric field induced at a point on the circumference of the circle, (c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.$\frac{di}{dt}=0.01 A/s \\$For 2s$\frac{di}{dt}$=0.02 A/s$\\$n=2000 turn/m, R=6.0cm=0.06m$\\$r=1cm=0.01m$\\a) \phi=BA \\\Rightarrow \frac{d\phi}{dt}=\mu_{\circ}nA\frac{di}{dt} \\$=$4\pi \times 10^{-7} \times 2 \times 10^3 \times \pi \times 1 \times 10^{-4} \times 2 \times 10^{-2} $[$A=\pi \times 1 \times 10^{-4}$]$\\$=$ 16\pi^2 \times 10^{-10} \omega \\$=$157.91 \times 10^{-10} \omega \\$=$1.6 \times 10^{-8} \omega \\$or,$\frac{d\phi}{dt} $for 1s=0.785$\omega \\b) \int E.dl=\frac{d\phi}{dt} \\\Rightarrow E\phi dl=\frac{d\phi}{dt} \\\Rightarrow E=\frac{0.785\times 10^{-8}}{2\pi \times 10^{-2}}=1.2\times 10^{-7} V/m \\$67. 66. An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor. Answer 67 None 68. 67. A magnetic flux of$8 \times 10^{-4} $weber is linked with each turn of a 200-turn coil when there is an electric current of 4 A in it. Calculate the self-inductance of the coil.$\frac{d\phi}{dt}=8\times 10^{-4} weber \\n=200, I=4A,E=-nL\frac{dI}{dt} \\$or,$\frac{-d\phi}{dt}=\frac{-LdI}{dt} \\or, L=n\frac{d\phi}{dt}=200 \times 8 \times 10^{-4}=2\times 10^{-2} H \\$69. 68. The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A/s. Find the emf induced in it. We know$ i=i_{\circ}(1-e^{\frac{-\tau}{r}} ) \\a) \frac{90}{100}i_{\circ}=i_{\circ}(1-e^{-t/r}) \\\Rightarrow 0.9=1-e^{-t/r} \\\Rightarrow e^{-t/r}=0.1 \\$Taking$l$n from both the sides$\\l$n$e^{-t/r}=ln0.1 \Rightarrow -t=-2.3 \Rightarrow t/r=2.3 \\b) \frac{99}{100}i_{\circ}=i_{\circ}(1-e^{-t/r})\\\Rightarrow e^{-t/r}=0.01 \\$or,$l$ne$^{-t/r}=ln0.01 \\$or,-t/r=-4.6 or t/r=4.6$\\$70. 69. Find the value of$t/\tau$for which the current in an LR circuit builds up to (a) 90%, (b) 99% and (c) 99.9% of the steady-state value We know$ i=i_{\circ}(1-e^{\frac{-\tau}{r}} ) \\a) \frac{90}{100}i_{\circ}=i_{\circ}(1-e^{-t/r}) \\\Rightarrow 0.9=1-e^{-t/r} \\\Rightarrow e^{-t/r}=0.1 \\$Taking$l$n from both the sides$\\l$n$e^{-t/r}=ln0.1 \Rightarrow -t=-2.3 \Rightarrow t/r=2.3 \\b) \frac{99}{100}i_{\circ}=i_{\circ}(1-e^{-t/r})\\\Rightarrow e^{-t/r}=0.01 \\$or,$l$ne$^{-t/r}=ln0.01 \\$or,-t/r=-4.6 or t/r=4.6$\\$71. 70. An inductor-coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit.$i=2A,E=4V,L=1H \\R=\frac{E}{i}=\frac{4}{2}=2 \\i=\frac{L}{R}=\frac{1}{2}=0.5 \\$72. 71. A coil having inductance 2.0H and resistance 20$\Omega$is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant.$L=2.0H,R=20 \Omega, emf=4.0 V, t=0.20 S \\i_{\circ}=\frac{e}{R}=\frac{4}{20}, \tau=\frac{L}{R}=\frac{2}{20}=0.1 \\$a)$i=i_{\circ}(1-e^{-t/r})=\frac{4}{20}(1-e^{-0.2/0.1}) \\$=$0.17 A \\$b)$\frac{1}{2}Li^{2}=\frac{1}{2}\times2 \times {(0.17)}^2=0.0289=0.03 J \\$73. 72. A coil of resistance 40 CI is connected across a 4'0 V battery. 0'10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil. 72.$ R=40\Omega, E=4v, t=0.1,i=63mA \\i=i_{\circ}-(1-e^{tR/2}) \\\Rightarrow 63 \times 10^{-3}=4/40(1-e^{-0.1\times 40/L}) \\\Rightarrow 63 \times 10^{-3}=10^ {-1}(1-e^{-4/L}) \\\Rightarrow 63\times 10^{-2}=(1-e^{-4/l}) \\\Rightarrow1-0.63=e^{-4/L} \Rightarrow e^{-4/L} =0.37 \\\Rightarrow -4/L=ln(0.37)=-0.994 \\\Rightarrow L=\frac{-4}{-0.994}=4.024H=4H \\$74. 73. An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Q. resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the circuit is switched on.$L=5.0H,R=100\Omega, emf=2.0V \\t=20ms=20\times 10^{-3}s=2\times 10^{-2} s \\i_{\circ}=\frac{2}{100} now i=i_{\circ}(1-e^{-t/\tau}) \\\tau=\frac{L}{R}=\frac{5}{100} \Rightarrow i=\frac{2}{100}(1-e \frac{-2\times 10^{-2} \times 100}{5}) \\\Rightarrow i=\frac{2}{100}(1-e^{-2/5}) \\\Rightarrow 0.00659=0.0066. \\V=iR=0.0066 \times 100=0.66 V. \\$75. 74. The time constant of an LR circuit is 40 ms. The circuit is connected at t = 0 and the steady-state current is found to be 2.0 A. Find the current at (a) t = 10 ms (b) t = 20 ms, (c) t = 100 ms and (d) t = 1 s.$\tau=40 ms \\i_{\circ}=2 A \\$a)t=10ms$\\i=i_{\circ}(1-e^{-t/\tau})=2(1-e^{-10/40})=2(1-e^{-1/4}) \\$=$2(1-0.7788)=2(0.2211)^{A}=0.4422A=0.44A \\$b)$t=20ms \\i=i_{\circ}(1-e^{-t/r})=2(1-e^{-20/40})=2(1-e^{-1/2}) \\=2(1-0.606)=0.7869 A=0.79 A \\$c)t=100ms$\\i=i_{\circ}(1-e^{-t/\tau})=2(1-e^{-100/40})=2(1-e^{-10/4}) \\$=2(1-0.082)=1.835 A=1.8 A$\\$d)$t=1s \\i=i_{\circ}(1-e^{-t/\tau})=2(1-e^{-1/40 \times 10^{-3}})=2(1-e^{-10/40} ) \\$=$2(1-e^{-25})=2 \times 1=2A \\$76. 75. An LR circuit has L =1.0 H and R = 20$\Omega$. It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t = 100 ms, (b) t =-200 ms and (c) t= 1.0 s.$L=1.0H,R$=20$\Omega$,emf=2.0V$\\\tau =\frac{L}{R}=\frac{1}{20}=0.05 \\i_{\circ}=\frac{e}{R}=\frac{2}{20}=0.1A \\i=i_{\circ}(1-e^{-t})=i_{\circ}-i_{\circ}e^{-t} \\\Rightarrow \frac{di}{dt}=\frac{di_{\circ}}{dt}(i_{\circ}x-1/\tau \times e^{-t/\tau})=i_{\circ}/\tau e^{-t/\tau} \\$. So,$\\$a)$ t=100ms \Rightarrow \frac{di}{dt}=\frac{0.1}{0.05} \times e^{-0.1/0.05}=0.27 A. \\$b)$t=200ms \Rightarrow \frac{di}{dt}=\frac{0.1}{0.05} \times e^{-0.2/0.05}=0.0366 A \\$c)$t=1s \Rightarrow \frac{di}{dt}=\frac{0.1}{0.05} \times e^{-1/0.05}=4 \times 10^{-9} A \\$77. 76. What are the values of the self-induced emf in the circuit of the previous problem at the times indicated therein ? a) For first case at t=100ms$\\\frac{di}{dt}=0.27 \\$Induced emf=$L\frac{di}{dt}=1 \times 0.27=0.27 V \\$b) For the second case at t=200 ms$\\\frac{di}{dt}=0.036 \\$Induced emf=$L\frac{di}{dt}=1 \times 0.036=0.036V \\$c)For the third case at t=1 s$\\\frac{di}{dt}=4.1 \times 10^{-9} V \\$Induced emf=$L\frac{di}{dt}=4.1\times 10^{-9} V \\$78. 77. An inductor-coil of inductance 20 mH having resistance 10$\Omega$is joined to an ideal battery of emf 5'0 V. Find the rate of change of the induced emf at t= 0, (b) t = 10 ms and (c) t = 1.0 s. L=20 mH; e=5.0V,R=10$\Omega\\\tau =\frac{L}{R}=\frac{20\times 10^{-3}}{10}, i_{\circ}=\frac{5}{10} \\i=i_{\circ}{(1-e^{-t/\tau})}^2 \\\Rightarrow iR=i_{\circ}R-i_{\circ}Re^{-t/\tau^2} \\$a)$10\times \frac{di}{dt}=\frac{d}{dt}i_{\circ}R + 10 \times \frac{5}{10} \times \frac{10}{20\times 10^{-3}} \times e^{-0\times 10/2 \times 10^{-2} } \\$=$\frac{5}{2} \times 10^{-3} \times 1=\frac{5000}{2}=2500=2.5\times 10^{-3} V/s. \\$b)$\frac{Rdi}{dt}=R\times i_{\circ} \times \frac{1}{\tau} \times e^{-t/\tau} \\t=10ms=10 \times 10^{-3} S \\\frac{dE}{dt}=10\times \frac{5}{10} \times \frac{10}{20\times 10^{-3}} \times e^{-0.01\times 10/2\times 10^{-2}} \\$=16.844=17V/'$\\$c) For t=1s$\\\frac{dE}{dt}=\frac{Rdi}{dt}=\frac{5}{2} 10^3 \times e^{10/2\times 10^{-2}}=0.00 V/s$.$\\$79. 78. An LR circuit contains an inductor of 500 mH, a resistor of 25.0$\Omega$and an emf of 5 00 V in series. Find the potential difference across the resistor at t - (a) 20.0 ms, (b) 100 ms and (c) l.00 s.$L=500mH,R=25\Omega,E=5V \\$a) t= 20ms$\\i=i_{\circ}(1-e^{--tR/L})=\frac{E}{R}(1-E^{-tR/L}) \\\frac{5}{25}(1-e^{-20\times 10^{-3} \times 25/100 \times 10^{-3}})=\frac{1}{5}(1-e^{-1}) \\\frac{1}{5}(1-0.3678)=0.1264 \\$Potential difference iR=0.1264 \times 25=3.1606 V=3.16 V.$\\$b) t=100ms$\\i=i_{\circ}(1-e^{-tR/L})=\frac{E}{R}(1-E^{-tR/L}) \\$=$\frac{5}{25}(1-e^{-100\times 10^{-3} \times 25/100 \times 10^{-3}})=\frac{1}{5}(1-e^{-5}) \\$=$\frac{1}{5}(1-0.0067)=0.19864 \\$Potential difference =$ iR =0.19864 \times 25=4.9665=4.97 V. \\$c)t=1 sec$\\i=i_{\circ}(1-e^{-tR/L})=\frac{E}{R}(1-E^{-tR/L}) \\=\frac{5}{25}(1-e^{1-e^{1\times 25/100 \times 10^{-3} } } )=\frac{1}{5}(1-e^{-50}) \\$=$\frac{1}{5} \times 1=1/5 A \\$Potential difference =$iR=(1/5 \times 25) V=5 V.\\$80. 79. An inductor-coil of resistance 10$\Omega$and inductance 120 mH is connected across a battery of emf 6 V and internal resistance 2$\Omega$. Find the charge which flows through the inductor in (a) 10 ms, (b) 20 ms and (c) 100 ms after the connections are made.$L=120 mH=0.120 H \\R=10\Omega, emf=6,r=2 \\i=i_{\circ}(1-e^{-t/\tau}) \\$Now, dQ=idt$\\$=$i_{\circ}(1-e^{-t/\tau}) dt \\Q=\int_{}^{}dQ=\int_{0}^{1}i_{\circ}(1-e^{-t/\tau}) dt \\$=$i_{\circ}[\int_{0}^{t} dt-\int_{0}^{1}e^{-t/\tau} dt]=i_{\circ}[t-(-\tau)\int_{0}^{1}e^{-t/\tau}dt] \\$=$i_{\circ}[t+\tau(e^{-t/\tau})]=i_{\circ}[t+\tau e^{-t/\tau} \tau] \\$Now,$ i_{\circ}=\frac{6}{10+2}=\frac{6}{12}=0.5A \\\tau=\frac{L}{R}=\frac{0.120}{12}=0.01 \\$a) t=0.01 s$\\$So,$Q=0.5[0.01+0.01e^{-0.01/0.01}-0.01] \\=0.00183=1.8 \times 10^{-3} C=1.8 mC \\$b)$t=20ms= 2\times 10^{-2}=0.02 s \\$So,$Q=0.5[0.02+0.01e^{-0.02/0.01}-0.01] =0.005676=5.6 \times 10^{-3} C=5.6mC \\$c)t=100ms=0.1 s$\\$So,$ Q=0.5[0.1+0.01e^{-0.1/0.01}-0.01] \\$=$0.045C=45mC \\$81. 80. An inductor-coil of inductance 17 mH is constructed from a copper wire of length 100 m and cross-sectional area$1 mm^2$. Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper = 1.7 x 10 8$\Omega$-m.$L=17mH,l=100m,A=1mm^{2}=1\times 10^{-6}m^2,f_{cu}=1.7 \times 10^{-8} \Omega m \\R=\frac{f_{cu}l}{A}=\frac{1.7 \times 10^{-8} \times 100}{1 \times 10^{-6}}=1.7\Omega \\i=\frac{L}{R}=\frac{0.17 \times 10^{-8}}{1.7}=10^{-2}sec=10 m sec \\$. 82. 81. An LR circuit having a time constant of 50 ms is connected with an ideal battery of emf$\epsilon$. Find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value and (c) the magnetic field energy stored in the circuit reaches half its maximum value.$\tau=L/R=50ms=0.05 \\$a)$\frac{i_{\circ}}{2}=i_{\circ}(1-e^{-t/0.06}) \\\Rightarrow \frac{1}{2}=1-e^{-t/0.05}=e^{-t/0.05}=\frac{1}{2} \\\Rightarrow \frac{1}{2}=1-e^{-t/0.05}=e^{-t/0.05}=\frac{1}{2} \\\Rightarrow ln=e^{-t/0.05}=ln^{1/2} \\\Rightarrow t=0.05\times 0.693=0.3465=34.6 ms=35 ms \\$b)$P=i^{2}R=\frac{E^2}{R}(1-E^{-tR/L})^2 \\$Maximum power=$\frac{E^2}{R} \\$So,$\frac{E^2}{2R}=\frac{E^2}{R}(1-e^{-tR/L})^2 \\\Rightarrow 1-e^{-tR/L}=\frac{1}{\sqrt{2}}0.707 \\\Rightarrow e^{-tR/L}=0.293 \\\Rightarrow \frac{tR}{L}=-In0.293=1.2275 \\\Rightarrow t=50 \times 1.2275ms=61.2 ms \\$83. 82. A coil having an inductance L and a resistance R is connected to a battery of emf$\epsilon$. Find the time taken for the magnetic energy stored in the circuit to change from one fourth of the steady-state value to half of the steady-state value. Maximum current=$\frac{E}{R} \\$In steady state magnetic field energy stored =$\frac{1}{2}L\frac{E^2}{R^2} \\$The fourth of steady state energy=$\frac{1}{8}L\frac{E^2}{R^2} \\$One half of steady energy=$\frac{1}{4}L\frac{E^2}{R^2} \\\frac{1}{8}L\frac{E^2}{R^2}=\frac{1}{2}L\frac{E^2}{R^2}(1-e^{-t_1R/L)^2} \\\Rightarrow 1-e^{-t_1R/L}=\frac{1}{2}\\\$