Magnetic Effect of Current

First, let's consider the magnetic field created by the long wire carrying a current $I$ at a distance $b$. According to Ampere's Law, the magnetic field $B$ at a distance $b$ from a long straight wire is given by:

$B = \frac{{\mu_0 I}}{{2\pi b}}$

where $\mu_0$ is the permeability of free space.

Given that the square loop of side $2a$ is carrying the same current $I$ and is located in the xz-plane with its center at the origin, we can analyze the forces on each side of the loop. The sides of the loop parallel to the x-axis will experience forces due to the magnetic field from the long wire. Considering symmetry and the directions of forces, the net force on these sides will not contribute to the torque around the z-axis.

The contribution to the torque around the z-axis will predominantly come from the sides of the loop parallel to the y-axis. For these sides, the magnetic forces will be in opposite directions and will create a torque around the z-axis.

Let's calculate the forces on the sides parallel to the y-axis. For a current element $Idl$ in the presence of a magnetic field $B$, the force $dF$ is given by:

$dF = I dl \times B$

For the sides at $x = a$ and $x = -a$, the distances to the wire are $a + b$ and $a - b$, respectively.

The magnetic fields at these positions due to the long wire are:

For $x = a$:

$B_a = \frac{{\mu_0 I}}{{2 \pi (a+b)}}$

For $x = -a$:

$B_{-a} = \frac{{\mu_0 I}}{{2 \pi (a-b)}}$

Since $b \gg a$, we can approximate these fields using binomial expansion for small $\left(\frac{a}{b}\right)$:

$B_a \approx \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right)$

$B_{-a} \approx \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right)$

The forces on each side of the loop with length $2a$ are:

For $x = a$:

$F_a = I \cdot 2a \cdot B_a = I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right)$

For $x = -a$:

$F_{-a} = I \cdot 2a \cdot B_{-a} = I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right)$

The net torque $\tau$ around the z-axis is due to these forces, with lever arms $a$ and $-a$ respectively:

$\tau = 2a \left( F_a - F_{-a} \right)$

Substituting the expressions for $F_a$ and $F_{-a}$:

$\tau = 2a \left[ I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right) - I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right) \right]$

Simplifying this expression, we get:

$\tau = 2a \left[ 2a I \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left( -\frac{2a}{b} \right) \right]$

$\tau = -2a \cdot \frac{{4a^2 \mu_0 I^2}}{{2 \pi b^2}}$

The negative sign indicates the direction of the torque, but the magnitude is:

$\tau = \frac{{4a^3 \mu_0 I^2}}{{\pi b^2}}$

Since $a \ll b$, the approximate magnitude of the torque around the z-axis simplifies to:

$\tau = \frac{{2 \mu_0 I^2 a^2}}{{\pi b}}$

Therefore, the correct answer is:

Option A

$\frac{{2\mu_0 I^2 a^2}}{{\pi b}}$

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)$