# Magnetic Field

## Concept Of Physics

### H C Verma

1   An alpha particle is projected vertically upward with a speed of 3 0 x 10 4 km/s in a region where a magnetic field of magnitude l'O T exists in the direction south to north. Find the magnetic force that acts on the a-particle.

##### Solution :

1. q = 2 ×1.6 × 10$^{–19}$ C, $u$ = 3 × 10$^4$ km/s = 3 × 10$^7$ m/s B = 1 T, F = qB$u$ = 2 × 1.6 × 10$^{–19}$ × 3 × 10$^7$ × 1 = 9.6 10$^{–12}$ N. towards west.

2   An electron is projected horizontally with a kinetic energy of $10$ keV. A magnetic field of strength $1.0$ x $10$ $T$ exists in the vertically upward direction, $\\$(a) Will the electron deflect towards right or towards left of its motion ?$\\$ (b) Calculate the sideways deflection of the electron in travelling through $1$ $m$. Make appropriate approximations.

##### Solution :

By solving we get, s = $0.0148$ $\approx$ $1.5$ x $10^{-2}$ cm

$KE$ = $10$ Kev = $1.6$ × $10^{-15}$ J, B = $1$ × $10$ T $\\$ $(a)$ The electron will be deflected towards left$\\$ $(b)$ $(1/2)$ $mv^2$ = KE $\Rightarrow$ V = $\sqrt{\frac{KE\times 2}{m}}$ F = qVB & accln = $\frac{qVB}{m_e}$

= $\frac{qBx^2}{2m_e\sqrt{\frac{KE \times 2}{m}}}$ = $\frac{1}{2}$ x $\frac{1.6 \times 10^{-19} \times 1 \times 10^{-7} \times 1^{2}} {9.1 \times 10^{-31} \times \sqrt{\frac{1.6 \times 10^{-15} \times 2}{9.1 \times 10^{-31}}}}$

Applying s = ut + $(1/2)$ at$^2$ = $\frac{1}{2}$ x $\frac{qVB}{m_e}$ x $\frac{x^2}{V^2}$ = $\frac{qBx^2}{2m_eV}$

3   A magnetic field of $(4.0 x 10^{-3} \vec{k})$ T exerts a force of $(4.0 \vec{i} +30 \vec{j})$ x $10^{-10}$ N on a particle having a charge of $1.0$ x $10^{9}$ $C$ and going in the $X-Y$ plane. Find the velocity of the particle.

##### Solution :

$F_X$ = $quV_YB$ $\Rightarrow$ $V_Y$ =$\frac{F}{qB}$ = $\frac{4 \times 10^{-10}}{1 \times 10^{-9} \times 4 \times 10^{-3}}$ = $100$ m/s

B = $4$ × $10^{–3}$ T$(Kˆ )$

Velocity = $(–75 \hat{i} + 100 \hat{j} )$ m/s

Considering the motion along x-axis :–

$F_Y$ $qV_xB$ $\Rightarrow$ $V_x$ =$\frac{F}{qB}$ = $\frac{3 \times 10^{-10}}{1 \times 10^{-9} \times 4 \times 10^{-3}}$ = $75$ m/s

F = [$4$ $\hat{i}$ + $3$ $\hat{j}$ × $10{–10}$] N.$\\$ $F_X$ = $4$ × $10^{–10}$ N $\\$ $F_Y$ = $3$ × $10^{–10}$ N $\\$ Q = $1$ × $10^{–9}$ C. $\\$

4   An experimenter's diary reads as follows: "a charged particle is projected in a magnetic field of $(7.0 \vec{i} - 3.0 \vec{j})$ x $10^{-3}$ T. The acceleration of the particle is foundtobe (---$\vec{i} + 7.0\vec{j}$) x $10^{-6}$ m/s$^2$".Thenumbertothe left of $\vec{i}$ in the last expression was not readable. What can this number be ?

##### Solution :

$\vec{B}$ = ($7.0$ $i$ – $3.0$ $j$) × $10^{-3}$ T $\\$ $\vec{a}$ = acceleration = (---$i$ + $7j$) × $10^{-6}$ m/s$^2$ $\\$ Let the gap be x. $\\$ Since $\vec{B}$ and $\vec{a}$ are always perpendicular $\\$ $\vec{B}$ x $\vec{a}$ = 0 $\\$ $\Rightarrow$ ($7x$ × $10^{–3}$ × $10^{–6}$ – $3$ × $10^{–3}$ $7$ × $10^{–6}$) = $0$ $\\$ $\Rightarrow$ $7x$ – $21$ = $0$ $\Rightarrow$ $x$ = $3$

5   A $10$ $g$ bullet having a charge of $4.00$ $\mu$C is fired at a speed of $270$ m/s in a horizontal direction. A vertical magnetic field of $500$ $\mu$T exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through $100$ m. Make appropriate approximations.

##### Solution :

$m$ = $10$ $g$ = $10$ × $10^{–3}$ $kg$ $\\$ $q$ = $400$ $mc$ = $400$ × $10^{–6}$ C$\\$ $u$ = $270$ $m/s$,$\\$ $B$ = $500$ $\mu t$ = $500$ × $10^{–6}$ Tesla

Along – $X axis$ $100$ = $270$ × $t$ $\Rightarrow$ $t$ = $\frac{10}{27}$ $\\$ Along – $Z axis$ $s$ = $ut$ + $(1/2)$ $at^2$ $\\$ $\Rightarrow$ $s$ = $\frac{1}{2}$ x $54$ x $10^{-6}$ x $\frac{10}{27}$ x $\frac{10}{27}$ = $3.7$ x $10^{-6}$

Force on the particle = $quB$ = $4$ × $10^{–6}$ × $270$ × $500$ × $10^{–6}$ = $54$ x $10^{-8}$ $(K)$ $\\$ Acceleration on the particle = $54$ × $10^{–6}$ $m/s^{2}$ $(K)$ $\\$ Velocity along $\vec{i}$ and acceleration along $\vec{k}$ $\\$ along $x-axis$ the motion is uniform motion and along $y-axis$ it is accelerated motion.

6   When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$, it moves with an initial acceleration $3a_0$ towards west. Find the electric field and the maximum possible magnetic field in the room.

##### Solution :

$q_P$ = $e$, $mp$ = $m$, $F$ = $q_P$ × $E$ $\\$ or $ma_0$ = $eE$ or, $E$ = $ma_0$ towards west $\\$ The acceleration changes from $a_0$ to $3a_0$ $\\$ Hence net acceleration produced by magnetic field $\vec{B}$ is $2a_0$.$\\$ Force due to magnetic field $\\$ = $\vec{F_B}$ = $m$ × $2a_0$ = $e$ × $V_0$ × $B$ $\\$ $\Rightarrow$ $B$ = $\frac{2ma_0}{eV_0}$ downwords

7   Consider a $10$ $cm$ long portioifof a straight wire carrying a current of 10 A placed in a magnetic field of $0.1$ $T$ making an angle of $53°$ with the wire. What magnetic force does the wire experience ?

##### Solution :

l = $10$ $cm$ = $10$ × $10^{–3}$ $m$ = $10^{–1}$ $m$ $\\$ i = $10$ $A$, $B$ = $0.1$ $T$, $\theta$ = $53°$ $\\$

$F$ = iLB $Sin$ $\theta$ = $10$ × $10^{–1}$ × $0.1$ × $0.79$ = $0.0798$ $\approx$ $0.08$ $\\$ direction of F is along a direction $\bot$r to both l and B.

8   A current of $2$ A enters at the corner $d$ of a square frame $abcd$ of side $20$ $cm$ and leaves at the opposite corner $b$. A magnetic field $B$ = $0.1$ $T$ exists in the space in a direction perpendicular to the plane of the frame as shown in figure $(34-E1)$ Find the magnitude and direction of the magnetic forces on the four sides of the frame.

##### Solution :

$\vec{F}$ = ilB = $1$ × $0.20$ × $0.1$ = $0.02$ N $\\$ For $F$ = il × $B$ $\\$ So, For $\\$ $da$ & $cb$ $\rightarrow$ l x $B$ $sin$ $90^{0}$ towards left $\\$ Hence $\vec{F}$ $0.02$ N towards left $\\$ for $\\$ $dc$ & $ab$ $\vec{F}$ = $0.02$ N downwards

9   A magnetic field of strength $1.0$ $T$ is produced by a strong electromagnet in a cylindrical region of radius $4.0$ $cm$ as shown in figure $(34-E2)$. A wire, carrying a current of $2.0$ $A$, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.

##### Solution :

$F$ = ilB $Sin$ $\theta$ $\\$ = ilB $Sin$ $90°$$\\ = i 2RB\\ = 2 × (8 × 10^{–2}) × 1 = 16 × 10^{–2}$$\\$ = 0.16 N.

10   A wire of length I carries a current i along the X-axis. A magnetic field exists which is given as $\vec{B}$ - $B_0$ ($\vec{i}$ + $\vec{j}$ + $\vec{k}$) T. Find the magnitude of the magnetic force acting on the wire.

##### Solution :

Length = l, Current = l $\hat{i}$ $\\$ $\vec{B}$ - $B_0$ ($\hat{i}$ + $\hat{j}$ + $\hat{k}$) T = $B_0\hat{i}$ + $B_0\hat{j}$ + $B_0\hat{k}$ $\\$ $F$ = Il x $\vec{B}$ = Il$\hat{i}$ x $B_0\hat{i}$ + $B_0\hat{j}$ + $B_0\hat{k}$ $\\$ = Il $B_0\hat{i}$ x $\hat{i}$ + l $B_0\hat{i}$ x $\hat{j}$ + l $B_0\hat{i}$ x $\hat{k}$ = Il $B_0\hat{k}$ - Il $B_0\hat{j}$

or, $\mid \vec{F} \mid$ = $\sqrt{2I^2I^2B_0^2}$ = $\sqrt{2}I l B_0$

11   A current of $5.0$ $A$ exists in the circuit shown in figure $(34-E3)$. The wire $PQ$ has a length of $50$ $cm$ and the magnetic field in which it is immersed has a magnitude of $0.20$ $T$. Find the magnetic force acting on the wire $PQ$.

##### Solution :

$i$ = $5$ $A$, $l$ = $50$ $cm$ = $0.5$ $m$ $\\$ $B$ = $0.2$ $T$,$\\$ $F$ = ilB $Sin$ $\theta$ = ilB $Sin$ $90°$ = $5$ × $0.5$ × $0.2$ $\\$ = $0.05$ N $\\$ $(\hat{j})$

12   A circular loop of radius $a$, carrying a current $i$, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field $(figure 34-E4)$. The strength of the magnetic field at the periphery of the loop is $B$. Find the magnetic force on the wire.

##### Solution :

$l$ = $2\pi a$ $\\$ Magnetic field = $B$ radially outwards $\\$ Current $\Rightarrow$ ‘$i$’ $\\$ $F$ = $i$ $l$ × $B$ $\\$ = $i$ × ($2\pi a$ × $\vec{B}$ ) $\\$ = $2\pi ai$ $B$ perpendicular to the plane of the figure going inside.

13   A hypothetical magnetic field existing in a region is given by $\vec{B}$ = $B_0$$\vec{e_r}, where er denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the centre at (0, 0, d). Find the magnituae of the magnetic force acting on the loop. ##### Solution : \vec{B} = B_0$$\vec{e_r}$ $\\$ $\vec{e_r}$ = Unit vector along radial direction $\\$ $F$ = $i$( $\vec{I}$ x $\vec{B}$ ) = ilB $Sin$ $\theta$ $\\$ = $\frac{i(2\pi a)B_0 a}{\sqrt{a^2+d^2}}$ = $\frac{i2\pi a^2)B_0}{\sqrt{a^2+d^2}}$

14   A rectangular wire-loop of width $a$ is suspended from the insulated pan of a spring balance as shown in figure $(34-E5)$. A current $i$ exists in the anticlockwise direction in the loop. A magnetic field $B$ exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

##### Solution :

Current anticlockwise$\\$ Since the horizontal Forces have no effect.$\\$ Let us check the forces for current along AD & BC [Since there is no $\vec{B}$ ] $\\$ In AD, F = 0 $\\$ For BC $\\$ F = iaB upward $\\$ Current clockwise $\\$ Similarly, F = – iaB downwards $\\$ Hence change in force = change in tension $\\$ = iaB – (–iaB) = $2$ iaB

15   A current loop of arbitrary shape lies in a uniform magnetic field $B$. Show that the net magnetic force acting on the loop is zero.

##### Solution :

$F_1$ = Force on AD = $ilB$ inwards $\\$ $F_2$ = Force on BC = $ilB$ inwards $\\$ They cancel each other $\\$ $F_3$ = Force on CD = $ilB$ inwards $\\$ $F_4$ = Force on AB = $ilB$ inwards $\\$ They also cancel each other.$\\$ So the net force on the body is $0$.

16   Prove that the force acting on a current-carrying wire, joining two fixed points $a$ and $b$ in a uniform magnetic field, is independent of the shape of the wire.

##### Solution :

For force on a current carrying wire in an uniform magnetic field $\\$ We need, l $\rightarrow$ length of wire $\\$ i $\rightarrow$ Current $\\$ B $\rightarrow$ Magnitude of magnetic field$\\$ Since $\vec{F}$ = $ilB$ $\\$ Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire.

17   A semicircular wire of radius $5.0$ $cm$ carries a current of $5.0$ $A$. A magnetic field B of magnitude $0.50$ T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.

##### Solution :

Force on a semicircular wire $\\$ = 2iRB $\\$ = $2$ × $5$ × $0.05$ × $0.5$ $\\$ = $0.25$ N

18   A wire, carrying a current $i$, is kept in the X-Y plane along the curve $y$ - $A$ $sin$$(\frac{2\pi}{\lambda}x). A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x = 0 and x = \lambda ##### Solution : Here the displacement vector \vec{dI} = \lambda \\ So magnetic for i \rightarrow t\vec{dl} x \vec{B} = i × \lambda B 19 A rigid wire consists of a semicircular portion of radius R and two straight sections (figure 34-E6). The wire is partially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current i. ##### Solution : Force due to the wire AB and force due to wire CD are equal and opposite to each \\ other. Thus they cancel each other.\\ Net force is the force due to the semicircular loop = 2iRB 20 A straight, horizontal wire of mass 10 mg and length 1.0 m carries a current of 2.0 A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight? ##### Solution : Mass = 10 mg = 10^{–5} kg \\ Length = 1 m \\ I = 2 A, B = ? \\ Now, Mg = ilB \\ \Rightarrow B = \frac{mg}{il} = \frac{10^{-5}\times 9.8}{2 \times 1} = 4.9 x 10^{-5} T 21 Figure (34-E7) shows a rod PQ of length 20.0 cm and mass 200 g suspended through a fixed point O by two threads of lengths 20.0 cm each. A magnetic field of strength 0.500 T exists in the vicinity of the wire PQ as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ. \\ (a) Find the tension in the threads when the switch S is open,\\ (b) A current of 2.0 A is established when the switch S is closed.\\ Find the tension in the threads now. ##### Solution : (a) When switch S is open\\ 2T Cos 30° = mg \\ \Rightarrow T = \frac{mg}{2cos30^0} \\ \frac{200 \times \ 10^{-3} \times 9.8}{2\sqrt{(3/2)}} = 1.13 (b) When the switch is closed and a current passes through the circuit = 2 A Then \\ \Rightarrow 2T Cos 30° = mg + ilB \\ = 200 × 10^{–3} 9.8 + 2 × 0.2 × 0.5 = 1.96 + 0.2 = 2.16 \\ \Rightarrow 2T = \frac{2.16 \times 2}{\sqrt{3}} = 2.49 \\ \Rightarrow T = 2.49 = 1.245 \approx 1.25 22 Two metal strips, each of length I, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in figure (34-E8). A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is \mu.A current i is established when the switch S is closed at the instant t = 0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach ? ##### Solution : Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be the dist covered.\\ So, F × l = \mumg × x \\ \Rightarrow ibBl = \mumgx \\ \Rightarrow x = \frac{ibBl}{\mu mg} 23 A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4.90 cm (figure 34-E9). A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 \Omega, the wire PQ starts sliding on the rails. Find the coefficient of friction. ##### Solution : \mu$$R$ = $F$ $\\$ $\Rightarrow$ $\mu$ x $m$ $g$ = ilB $\\$ $\Rightarrow$ $\mu$ x $10$ $10^{-3}$ x $9.8$ = $\frac{6}{20}$ x $4.9$ x $10^{2}$ x$0.8$ $\\$ $\Rightarrow$ $\mu$ = $\frac{0.3 \times 0.8 \times 10^{-2}}{2 \times 10^{-2}}$ = $0.12$

24   A straight wire of length $I$ can slide on two parallel plastic rails kept in a horizontal plane with a separation $d$. The coefficient of friction between the wire and the rails is $\mu$. If the wire carries a current $i$, what minimum magnetic field should exist in the space in order to slide the wire on the rails.

27   The magnetic field existing in a region is given by $\\$ $\vec{B}$ = $B$$\big(1+\frac{x}{i}\big) \vec{k}.\\ A square loop of edge l and carrying a current i, is placed with its edges parallel to the X-Y axes. Find the magnitude of the net magnetic force experienced by the loop. ##### Solution : \vec{B} = B$$\big(1+\frac{x}{i}\big)$ $\vec{k}$ $\\$ $f_1$ = force on AB = i$B_0$[1 + 0]l = i$B_0$l $\\$ $f_2$ = force on CD = i$B_0$[1 + 0]l = i$B_0$l $\\$ $f_3$ = force on AD = i$B_0$[1 + 0/1]l = i$B_0$l $\\$ $f_4$ = force on AB = i$B_0$[1 + 1/1]l = 2i$B_0$l $\\$ Net horizontal force = $F_1$ – $F_2$ = $0$ $\\$ Net vertical force = $F_4$ – $F_3$ = i$B_0$l $\\$

28   A conducting wire of length $I$, lying normal to a magnetic field $B$, moves with a velocity $v$ as shown in figure $(34-E11)$. $\\$ (a) Find the average magnetic force on a free electron of the wire,$\\$(b) Due to this magnetic force, electrons concentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops, $\\$ (c) What potential difference is developed between the ends of the wire ?

##### Solution :

$(a)$ Velocity of electron = $v$ $\\$ Magnetic force on electron $\\$ $F$ = $evB$ $\\$ $(b)$ $F$ = $qE$; $F$ = $evB$ or, $\vec{E}$ = $vB$ $\\$ $(c)$ $E$ = $\frac{dr}{dV}$ = $\frac{v}{B}$ $\Rightarrow$ $V$ = $lE$ = $lvB$

29   A current i is passed through a silver strip of width $d$ and area ot cross-section $A$. The number of free electron per unit volume is $n$.$\\$ (a) Find the drift velocity u of the electrons,$\\$ (b) If a magnetic field $B$ exists in the region as shown in figure $(34-E12)$, what is the average magnetic force on the free electrons ?$\\$ (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons, $\\$(d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation $?$ The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called $Hall$ effect.

##### Solution :

$(a)$ $i$ = $V_0$$n$$Ae$ $\Rightarrow$ $V_0$ = $\frac{i}{nae}$ $\\$ $(b)$ $F$ = $\frac{iBl}{an}$ = $\frac{iB}{an}$ (upwards) $\\$ $(c)$ Let the electric field be $E$ $\\$ $Ee$ = $\frac{iB}{an}$ $\Rightarrow$ $E$ = $\frac{iB}{Aenl}$ $\\$ $(d)$ $\frac{dv}{dr}$ = $E$ $\Rightarrow$ $dV$ = $Edrl$ $\\$ = $E$ x $d$ = $\frac{iB}{Aen}$ $d$

30   A particle having a charge of $2.0$ x $10$$^8 C and a mass of 2.0 x 10$$^{16}$ g is projected with a speed of $2.0$ x $10^3$ $m/s$. in a region having a uniform magnetic field of $0.10$ $T$. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period,

##### Solution :

$q$ = $2.0$ × $10^{–8}$ C $vec{B}$ = $0.10$ $T$ $\\$ $m$ = $2.0$ × $10^{–10}$ g = $2$ × $10^{–13}$ g $\\$ $v$ = $2.0$ × $103$ $m/ '$

$R$ = $\frac{mv}{q|B}$ $\frac{2 \times 10^{-13} \times 2 \times 10^{3}}{ 2 \times 10^{-8} \times 10^{-1} }$ = $0.2$ $m$ = $20$ cm$\\$ $T$ = $\frac{2\pi m}{qB}$ $\frac{2 \times 3.14 \times 2 \times 10^{3}}{ 2 \times 10^{-8} \times 10^{-1} }$ = $6.28$ x $10^{-4}$s

31   A proton describes a circle of radius $1$ cm in a magnetic field of strength $0.10$ $T$. What would be the radius of the circle described by an $\alpha$-particle moving with the same speed in the same magnetic field ?

##### Solution :

$r$ = $\frac{mv}{qB}$ $\\$ $0.01$ = $\frac{mv}{e0.1}$ ....(1) $\\$ $r$ = $\frac{4m\times V}{2e \times 0.1}$ .....(2)$\\$ $(2)$ $\div$ $(1)$ $\\$ $\Rightarrow$ $\frac{r}{0.01}$ = $\frac{4mVe\times 0.1}{2e \times 0.1 \times mv}$ = $\frac{4}{2}$ = $2$ $\Rightarrow$ $r$ = $0.02$ $m$ = $2$ cm.

32   An electron having a kinetic energy of $100$ $eV$ circulates in a path of radius $10$ $cm$ in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

##### Solution :

$KE$ = $100ev$ = $1.6$ × $10^{–17}$ $J$ $\\$ $(1/2)$ × $9.1$ × $10^{–31}$ × $V^2$ = $1.6$ × $10^{–17}$ $J$ $\\$ $\Rightarrow$ $V^2$ = $\frac{1.6$ \times $10^{–17 \times 2}}{9.1 \times 10^{-31}}$ = $0.35$ x $10^{14}$

$T$ = $\frac{2\pi m}{qB}$ = $\frac{2 \times 3.14 \times 9.1\times 10^{-31}}{1.6 \times 10^{-19} \times 3.4 \times 10^{-4}}$ $\\$ No. of Cycles per Second $f$ = $\frac{1}{T}$ $\\$ = $\frac{1.6 \times 3.4}{2 \times 3.14 \times 9.1}$ x $\frac{10^{-19} \times 10^{-4}}{10^{-31}}$ = $0.0951$ x $10^8$ $\approx$ $9.51$ x $10^6$$\\ Note: \therefore Puttig \vec{B} 3.361 × 10^{–4} T We get f = 9.4 × 10^6 or, V = 0.591 × 10^7 m/s \\ Now r = \frac{mv}{qB} \Rightarrow \frac{9.1 \times 10^{-13} \times 0.591 \times 10^{7} }{1.6 \times 10^{-19} \times B} = \frac{10}{100} \\ \Rightarrow B = \frac{9.1 \times 0.591}{1.6} x \frac{10^{-23}}{10^{-19}} = 3.3613 x 10^{-4} T \approx 3.4 x 10^{-4} T 33 Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance I in front of the accelerator. Find the magnetic field. ##### Solution : Radius = l, K.E = K \\ L = \frac{mV}{qB} \Rightarrow l = \frac{\sqrt{2mk}}{qB} \\ \Rightarrow B = \frac{\sqrt{2mk}}{ql} 34 A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 x 10^6 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it. ##### Solution : V = 12 kv E = \frac{V}{l} Now,F = qe =\frac{qv}{l} or, a = \frac{F}{m} = \frac{qVl}{ml} \\ v = 1 x10^6 m/s \\ or V = \sqrt{2 \times \frac{qV}{ml} \times l} = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^3} \\ or 1 x 10^6 = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^3} \\ \Rightarrow 10^{12} = 24 × 10^3 × \frac{q}{m} \Rightarrow \frac{m}{q} = \frac{24 \times 10^3 }{10^12} = 24 x 10^{-9} \\ r = \frac{mV}{qB} = \frac{24 \times 10^{-9} \times 1 \times 10^6}{2 \times 10^{-1}} = 12 x 10^{-2} m = 12 cm \\ 35 Doubly-ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 1.0 T. Find \\ (a) the force acting on an ion, \\ (b) the radius of the circle in which it circulates and \\ (c) the time taken by an ion to complete the circle. ##### Solution : V = 10 Km/$$^'$ = $10^4$ $m/s$ $\\$ $B$ = $1$ $T$, $q$ = $2e$. $\\$ $(a)$ $F$ = $qVB$ = $2$ × $1.6$ × $10^{–19}$ × $10^4$ × $1$ = $3.2$ × $10^{–15}$ $N$ $\\$ $(b)$ $r$ = $\frac{mV}{qB}$ = $\frac{4 \times 1.6 \times 10^{-27} \times 10^4}{ 2 \times 1.6 \times 10^{-19} \times 1}$ = $2$ x $\frac{10^{-23}}{10^{-19}}$ = $2$ x $10^{-4}$ $m$ $\\$ $(c)$ Time taken = $\frac{2\pi r}{V}$ = $\frac{2\pi mv}{qB \times V}$ $\frac{2 \pi \times 4 1.6 \times 10^{-27}}{2 \times 1.6 \times10^{-19} \times 1}$

= $4\pi$× $10^{–8}$ = $4$ × $3.14$ × $10{–8}$ = $12.56$ × $10^{–8}$ = $1.256$ × $10^{–7}$ $sec$.

36   A proton is projected with a velocity of $3 \times 10^6$ $m/s$ perpendicular to a uniform magnetic field of $0.6$ $T$. Find the acceleration of the proton.

##### Solution :

$u$ = $3$ × $10^6$ $m/s$, $B$ = $0.6$ $T$, $m$ = $1.67$ × $10^{–27}$ $kg$ $\\$ $F$ = $quB$ , $qP$ = $1.6$ × $10^{–19}$ $C$ $\\$ or, $vec{a}$ = $\frac{F}{m}$ = $\frac{quB}{m}$ $\\$ =$\frac{1.6 \times 10^{-19} \times 3 \times 10^{6} \times 10^{-1}}{1.67 \times 10^{-27}}$ $\\$ = $17.245$ × $10^13$ = $1.724$ × $10^4$ $m/s^2$

37   (a) An electron moves along a circle of radius $1$ $m$ in a perpendicular magnetic field of strength $0.50$ $T$. What would be its speed ? Is it reasonable ? $\\$ (b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed ?

$(a)$ $R$ = $1$ $n$ , $B$= $0.5$ $T$ , $r$ = $\frac{mu}{qB}$ $\\$ $\Rightarrow$ $1$ = $\frac{9.1 \times 10^{-31} \times u}{1.6 \times 10^{-19} \times 0.5}$ $\\$ $\Rightarrow$ $\frac{1.6 \times 0.5 \times 10^{-19}}{9.1 \times 10^{-31}}$ = $0.0879$ × $10^{10}$ $\approx$ $8.8$ × $10^{10}$ $m/s$$\\ No, it is not reasonable as it is more than the speed of light. \\ (b) \frac{mu}{qB} \\ \Rightarrow 1 = \frac{1.6 \times 10^{-27} \times v}{1.6 \times 10^{-19} \times 0.5} \\ \Rightarrow u = \frac{1.6 \times 10^{-19} \times 0.5}{1.6 \times 10^{-27}} = 0.5 x 10^8 = 5 x 10^7 m/s 38 A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure (34-E13).\\ (a) Find the radius of the circular arc it describes in the magnetic field,\\ (b) Find the angle subtended by the arc at the centre, \\ (c) How long does the particle stay inside the magnetic field ?\\ (d) Solve the three parts of the above problem if the charge q on the particle is negative. ##### Solution : (a) Radius of circular arc = \frac{mv}{qB} \\ (b) Since MA is tangent to are ABC, described by the particle.\\ Hence \angle{MAO} = 90° \\ Npw \angle{NAC} = 90° {\therefore NA is \bot r] \\ \therefore \angle{AOC} = \angle{OCA} = \theta [By geometry]\\ Then \angle{AOC} = 180 - (\theta - + \theta) = \pi - 20 \\ (c) Dist. Covered l = r\theta = \frac{mv}{qB}$$(\pi - 2 \theta )$ $\\$ $t$ = $\frac{l}{v}$ = $\frac{m}{qB}$$(\pi - 2 \theta ) \\ (d) If the charge ‘q’ on the particle is negative. Then \\ (i) Radius of Circular arc = \frac{mv}{qB} \\ (ii) In such a case the centre of the arc will lie with in the magnetic field, as seen \\ in the fig. Hence the angle subtended by the major arc = \pi + 2\theta \\ (iii) Similarly the time taken by the particle to cover the same path = \frac{m}{qB} (\pi + 2 \theta) 39 A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation (figure 34-E14) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than \\ (a) \frac{mv}{qB} \\ (b) \frac{mv}{2qB} \\ (c) \frac{2mv}{qB} ##### Solution : Mass of the particle = m , Charge = q , Width = d \\ (a) if d = \frac{mv}{qB} \\ The d is equal to radius. \theta is the angle between the \\ radius and tangent which is equal to \pi / 2 (As shown in the figure) \\ (b) if \approx \frac{mv}{2qB} distance travelled = (1/2) of radius \\ Along x-directions d = V_X t [Since acceleration in this direction is 0. Force acts along \hat{j} directions] t = \frac{d}{V_X} ....(1) \\ V_Y = u_Y + a_Y t = \frac{0 + qu_x Bt}{m} = \frac{qu_xBt}{m} \\ From (1) putting the value of t, V_Y = \frac{qu_xBd}{mV_x} \\ Tan$$\theta$ = $\frac{V_Y}{V_X}$ = $\frac{qBd}{mV_x}$ = $\frac{qBmV_X}{2qBmV_x}$ = $\frac{1}{2}$ $\\$ $\Rightarrow$ $\theta$ = $\tan^{-1}$ $(\frac{1}{2})$ = $26.4$ $\approx$ $30^0$ = $\frac{\pi}{6}$ $\\$ $(c)$ $d$ $\approx$ $\frac{2mu}{qB}$ $\\$ Looking into the figure, the angle between the initial direction and final direction of velocity is $\pi$.

40   A narrow beam of singly-charged carbon ions, moving at a constant velocity of $6.0$ x $10$ $m/s$, is sent perpendicularly in a rectangular region having uniform magnetic field $B$ = $0.5$ $T$ $(figure 34-E15)$. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being $3.0$ $cm$ and $.5$ $cm$. Identify the isotopes present in the ion beam. Take the mass of an ion = $A(1.6 x 10)$ $kg$, where A is the mass number.

$u$ = $6$ × $10^4$ $m/s$, $B$ = $0.5$ $T$, $r1$ = $3/2$ = $1.5$ $cm$, $r2$ = $3.5/2$ $cm$ $\\$ $r_1$ = $\frac{mV}{qB}$ = $\frac{A \times (1.6 \times 10^{-27})\times 6 \times 10^4}{1.6 \times 10^{-19} \times 0.5}$ $\\$ $\Rightarrow$ $1.5$ = A x $12$ x $10^{-4}$$\\ \Rightarrow A = \frac{1.5}{2 \times 10^{-4}} = \frac{15000}{12} \\ r_2 = \frac{mu}{qB} \Rightarrow \frac{3.5}{2} = \frac{A^{'} \times (1.6 \times 10^{-27})\times 6 \times 10^4}{1.6 \times 10^{-19} \times 0.5} \\ \Rightarrow A^{'} = \frac{3.5 \times 0.5 \times 10^{-19}}{2 \times 6 \times 10^4 \times 10^{-27}} = \frac{3.5 \times 0.5 \times 10^4}{12} \\ \frac{A}{A^{'}} = \frac{1.5}{12 \times 10^{-4}} \frac{12 \times 10^{-4}}{3.5 \times 0.5} = \frac{6}{7} \\ Taking common ration = 2 (For Carbon). The isotopes used are C^{12} and C^{14} 41 Fe * ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A (1.6 \times 10^{-3})kg where A is the mass number. ##### Solution : V = 500 V B = 20 mT = (2 \times 10^{–3}) T r_1 = \frac{m_1 \sqrt{1000 \times q_1}}{q_1\sqrt{m_1B}} = \frac{\sqrt{m_1} \sqrt{1000}} {\sqrt{q_1}B} = \frac{\sqrt{57 \times 1.6 \times 10^{-27} \times 10^3}}{1.6 \times 10^{-19} \times 2 \times 10^{-3}} = 1.19 x 10^{-2} m = 119 cm \\ r_1 = \frac{m_2 \sqrt{1000 \times q_2}}{q_2\sqrt{m_2 B}} = \frac{\sqrt{m_2} \sqrt{1000}} {\sqrt{q_2}B} = \frac{\sqrt{1000 \times 58 \times 1.6 \times 10^{-27}}}{1.6 \times 10^{-19} \times 20 \times 10^{-3}} = 1.20 x 10^{-2} m = 120 cm E = \frac{V}{d} \frac{500}{d} = \Rightarrow F \frac{q500}{d} \Rightarrow a \frac{q500}{d} \\ \Rightarrow u^2 = 2ad = 2 x \frac{q500}{dm} x d \Rightarrow u^2 = \frac{1000 \times q}{m} \Rightarrow u = \sqrt{\frac{1000 \times q}{m}} \\ 42 A narrow beam of singly-charged potassium ions of kinetic energy 32 ke V is injected into a region of width 1.00 cm having a magnetic field of strength 0.500 T as shown in figure (34-E16). The ions are collected at a screen 95.5 cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A (1.6 \times 10^{-27}) kg where A is the mass number. ##### Solution : V(in vertical direction) = at = 5193.535216 × 10^8 × 24 × 10^{–9} = 12464.48452 m/s^2 t = \frac{0.01}{40.50957468 \times 10^5} = 24 × 10^{–19} sec. [Time taken to cross the magnetic field] Net displacement from line = 0.0001495738143 + 0.0293912545 = 0.0295408283143 m \\ For K – 41 : (1/2) × 41 × 1.6 × 10^{–27} v = 32 × 10^3 1.6 × 10^{–19} \Rightarrow v = 39.50918387 m/s. For K - 39 : m = 39 \times 1.6 \times 10^{-27} kg, B = 5 x 10^{-1} T, q = 1.6 x 10^{-19}$$C$, $K.E$ = $32$ $Kev$.

= $\frac{1.6 \times 10^{-19} \times 4.050957468 \times 10^5 \times 0.5}{ 39 \times 1.6 \times 10{-27}}$ = $5193.535216$ × $10^8$ $m/s^2$

Time gap = $2442$ × $10^{–9}$ – $25$ × $10^{–9}$ = $2417$ × $10^{–9}$ $\\$ Distance moved vertically = $12045.48289$ × $2417$ × $10^{–9}$ = $0.02911393215$ $\\$ Now, $V^2$ = $2as$ $\Rightarrow$ $(12045.48289)^2$ = $2$ × $4818.193151$ × $S$ $\Rightarrow$ $S$ = $0.0001505685363$ $m$ $\\$ Net distance travelled = $0.0001505685363$ + $0.02911393215$ = $0.0292645006862$ $\\$ Net gap between $K$ – $39$ and $K$ – $41$ = $0.0295408283143$ – $0.0292645006862$ $\\$ = $0.0001763276281$ $m$ $\approx$ $0.176$ $mm$

Through out the emotion the horizontal velocity remains constant .

Time gap = $2383$ × $10^{–9}$ – $24$ × $10^{–9}$ = $2358$ × $10^{–9}$ $sec$. $\\$ Distance moved vertically (in the time) = $12464.48452$ × $2358$ × $10^{–9}$ = $0.0293912545$ $m$ $\\$ $V^2$ = $2as$ $\Rightarrow$ $(12464.48452)^2$ = $2$ × $5193.535216$ × $10^8$ × $S$ $\Rightarrow$ $S$ = $0.1495738143$ × $10^{–3}$ $m$.

Accln. In the region having magnetic field = $\frac{qvB}{m}$

$a$ = $\frac{qvB}{m}$ = $\frac{1.6 \times10^{19} \times 395091.8387 \times 0.5}{41 \times 1.6 \times 10^{-27}}$ = $4818.193154$ × $10^8$ $m/s^2$ $\\$ $t$ = (time taken for coming outside from magnetic field) = $\frac{0.01}{39501.8387}$ = $25 \times 10^{–9}$ $sec$. $\\$ $V$ = at (Vertical velocity) = $4818.193154$ × $10^8$ × $10^8$ $25$ × $10^{–9}$ = $12045.48289$ $m/s$. $\\$ (Time total to reach the screen) = $\frac{0.965}{395091.8387}$ = $0.000002442$

##### Solution :

Velocity will be along x – z plane $\\$ $\vec{B}$ = - $B_{0}$$\vec j \vec E = E_0 \vec k \\ F = q (\vec E + \vec V x \vec B = q = [E_0$$\hat{k}$ + ($u_x$ $\hat i$ + $u_x$ $\hat k$)(-$B_0$$\hat{j})] = (qE_0)$$\hat k$ = $(u_x B_0)$$\hat{k} + (u_z B_0)$$\hat{i}$ $\\$ $F_Z$ = ($qE_0 - u_XB_0$) $\\$ Since $u_x$ = $0$ , $F_Z$ = $qE_0$ $\Rightarrow$ $a_z$ = $\frac{qE_0}{m}$, So, $V^2$ = u$^2$ + $2as$ $\Rightarrow$ $v^2$ = 2$\frac{qE_0}{m}$$Z[distance along Z diretion be z] \\ \Rightarrow V = \sqrt{\frac{2qE_0Z}{m}} 52 An electron is emitted with negligible speed from the negative plate of a parallel plate capacitor charged to a potential difference V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure (34-E20). Show that the electron will fail to strike the upper plate if \\ d > \big(\frac{2m_pV}{eB^{2}_0}\big)$$^{1/2}$

##### Solution :

The force experienced first is due to the electric field due to the capacitor $\\$ $E$ = $\frac{V}{d}$ $F$ = $eE$ $\\$ $a$ = $\frac{eE}{m_e}$ [Where e $\rightarrow$ charge of electron $m_e$ $\rightarrow$ mass of electron] $\\$ $v^2$ = $u^2$ + $2as$ $\Rightarrow$ $v^2$ = $2$ x $\frac{eE}{m_e}$ x $d$ = $\frac{2 \times e \times V \times d}{dm_e}$ $\\$ or $v$ = $\sqrt{\frac{2eV}{m_e}}$ $\\$ Now, The electron will fail to strike the upper plate only when d is greater than radius of the are thus formed. $\\$ or, $d$ > $\frac{m_e \times \sqrt{\frac{2eV}{m_e}} }{eB}$ $\Rightarrow$ $d$ > $\frac{\sqrt{2m_eV}}{eB^2}$

53   A rectangular coil of $100$ turns has length $5$ $cm$ and width $4$ $cm$. It is placed with its plane parallel to a uniform magnetic field and a current of $2$ $A$ is sent through the coil. Find the magnitude of the magnetic field $B$, if the torque acting on the coil is $0.2$ $N-m$.