 # Electromagnetic Induction

## Concept Of Physics

### 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

##### Solution :

$\int_{}^{} E.dl =MLT^{-3}l^{-1}\times L=ML^{2}l^{-1}T^{-3}$ $\\$ (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

##### Solution :

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

##### Solution :

$\phi =at^2+bt+c$ $\\$ (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.

##### Solution :

$\phi_2=B.A=0.01\times2\times10^{-3}=2\times 10^{-5} m$ $\\$ $\phi_1=0$ $\\$ $e=-\frac{d\phi}{dt}=\frac{-2\times10^{-5}}{10\times10^{-3}}=-2mV$ $\\$ $\phi_3=B.A=0.03\times2\times 10^{-3}=6\times 10^{-5}$ $\\$ $d\phi = -4\times 10^{-5}$ $\\$ $e=-\frac{d\phi}{dt}=-4mV$ $\\$ $\phi_4=B.A=0.01\times2\times10^{-3}=2\times 10{^-5}$ $\\$ $d\phi=-4\times10^{-5}$ $\\$ 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.

##### Solution :

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.

##### Solution :

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.

##### Solution :

(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.

##### Solution :

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.

##### Solution :

$A=5cm^2=5\times10^{-4}m^2$ $\\$ $B=B_0sin\omega t=0.2sin(300)t$ $\\$ $\theta=60^{\circ}$ $\\$ 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 ?

##### Solution :

$\vec{B}=0.10T$ $\\$ 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.

##### Solution :

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.

##### Solution :

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.

##### Solution :

$r=2cm =2\times10^{-2} m$ $\\$ 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.

##### Solution :

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?

##### Solution :

$\phi_1=0.35 weber, \phi_2=0.85 weber$ $\\$ $D\phi=\phi_2-\phi_1=(0.85-0.35)weber=0.5 weber$ $\\$ 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.

##### Solution :

$i=v(B\times I)$ $\\$ =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.

##### Solution :

$u=1cm/',B=0.6T$ $\\$ 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.

##### Solution :

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.

##### Solution :

r=10cm, r=40cm $\\$ $\frac{dB}{dt}=0.010T/',\frac{d\phi}{dt}=\frac{dB}{dt}A$ $\\$ $E=\frac{d\phi}{dt}=\frac{dB}{dt} \times A=0.01(\frac{\pi \times r^2}{2})$ $\\$ $=\frac{0.01\times 3.14\times 0.01}{2}=\frac{3.14}{2}\times 10^{-4}=1.57\times 10^{-4}$ $\\$ $i=\frac{E}{R}=\frac{1.57\times10^{-4}}{4}=0.39\times 10^{-4}=3.9\times 10^{-5} A$ $\\$