Magnetic Effect of Current

There are two forces on slider.

Spring force = kx

where, k = spring constant.

As the slider is kept in a uniform magnetic field B = 0.1 T, hence it will experience a force, i.e.

Magnetic force = Bil

where, l = length of the strip.

Now, using

F_net = ma

We have,

$( - kx) + ( - Bil) = ma$

$ \Rightarrow - kx - Bil - ma = 0$

$ \Rightarrow - kx - {{{B^2}{l^2}} \over R}.\,v - ma = 0$ [$\because$ $i = {{Blv} \over R}$]

and acceleration, $a = {{{d^2}x} \over {d{t^2}}}$

Hence, the modified equation becomes

$ \Rightarrow {{m{d^2}x} \over {d{t^2}}} + {{{B^2}{l^2}} \over R}\left( {{{dx} \over {dt}}} \right) + kx = 0$

This is the equation of damped simple harmonic motion.

So, amplitude of oscillation varies with time as

$A = {A_e}^{ - {{{B^2}{l^2}} \over {2Rm}}.\,t}$

Now, when amplitude is ${{{A_0}} \over e}$, then

${{{A_0}} \over e} = {{{A_0}} \over {{e^{{{{B^2}{l^2}} \over {2Rm}}.\,t}}}}$ (as given)

$ \Rightarrow \left( {{{{B^2}{l^2}} \over {2Rm}}} \right)t = 1$ or $t = {{2Rm} \over {{B^2}{l^2}}}$

According to the question, magnetic field B = 0.1 T, mass of strip m = 50 $\times$ 10^$-$3 kg, resistance R = 10 $\Omega$, l = 10 cm = 10 $\times$ 10^$-$2 m

$\therefore$ $t = {{2Rm} \over {{B^2}{l^2}}} = {{2 \times 10 \times 50 \times {{10}^{ - 3}}} \over {{{(0.1)}^2} \times {{(10 \times {{10}^{ - 2}})}^2}}}$

$ = {1 \over {{{10}^{ - 4}}}} = 10000$ s

Given, spring constant, k = 0.5 Nm^$-$1

Also, time period of oscillation is

$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{{50 \times {{10}^{ - 3}}} \over {0.5}}} = {{2\pi } \over {\sqrt {10} }} \approx 2s$

So, number of oscillations is

$N = {t \over T} = {{10000} \over 2} = 5000$

According to the question, the situation can be drawn as

Let the current I is flowing in anti-clockwise direction, then the magnetic moment of the coil is

m = NIA

where, N = number of turns in coil

and A = area of each coil = $\pi$r².

Its direction is perpendicular to the area of coil and is along Y-axis.

Then, torque on the current coil is

$\tau = m \times B = mB\sin 90^\circ = NIAB$

$ = NI\pi {r^2}B(N - m)$

We know that magnetic field due to finite current carrying wire is

$B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {\theta _1} + \cos {\theta _2})$ ..... (1)

Given that side of triangle = 4.5 $\times$ 10^$-$2 m = a; current = 1A since the triangle is equilateral, angle of each side will be 60$^\circ$.

Now, $\tan \theta = {{Perpendicular} \over {Base}} \Rightarrow \tan 60^\circ = {{a/2} \over b}$

$ \Rightarrow b = {a \over {2\tan 60^\circ }} = {a \over {2\sqrt 3 }}$

Using equation (1), we get

$B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ $

$B = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 \times \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}$

$ \Rightarrow B = {{{{10}^{ - 7}} \times 6 \times 1} \over {4.5 \times {{10}^{ - 2}}}} = 1.33 \times {10^{ - 5}} \sim 2 \times {10^{ - 5}}$ Wb/m²

Given situation is shown in the figure

As we know,

$\overrightarrow F = q(\overrightarrow v \times \overrightarrow B )$

$\overrightarrow F = - {q_0}(\overrightarrow v \times \overrightarrow B )$

According to question, direction of current is parallel to the force acting on the electron. Hence, the motion of test charge is towards the wire.