Magnetic Effect of Current

$ \begin{aligned} B_{P_{1}} & =\frac{\mu_{0} l}{2 r} \\\\ B_{P_{2}} & =\frac{\mu_{0} l r^{2}}{2\left(r^{2}+r^{2}\right)^{3 / 2}}=\frac{\mu_{0} I}{2^{5 / 2} r} \end{aligned} $

$ \begin{aligned} & \therefore \quad \frac{B_{P_{1}}}{B_{P_{2}}}=\frac{\frac{\mu_{0} I}{2 r}}{\frac{\mu_{0} I}{2^{5 / 2} r}}=\frac{2 \sqrt{2}}{1} \end{aligned} $

Length of wire Y is very high compared to length of wire X. So we can assume length of wire Y is infinite compare to wire X.

Magnetic field due to wire y from 5 cm from wire towards wire X is,

$B = {{{\mu _0}{I_1}} \over {2\pi d}} = {{{\mu _0} \times 3} \over {2\pi \times 0.05}}$

And direction of B is outward.

Now force on wire X due to wire Y,

${F_{YX}} = {I_2}B\,l$

$ = 2 \times {{{\mu _0} \times 3} \over {2\pi \times 0.05}} \times 5$

$ = 1.2 \times {10^{ - 5}}\,N$

Current flowing in both the wire in same direction so both wire will attract each other with equal force.

$\therefore$ ${F_{YX}} = {F_{XY}}$

$\therefore$ ${F_{XY}} = 1.2 \times {10^{ - 5}}\,N$, attractive force towards wire X.

$\overrightarrow F = i\overrightarrow l \times \overrightarrow B $

$ = ilB\sin 60^\circ $

$ = 10 \times {5 \over {100}} \times 0.5 \times {{\sqrt 3 } \over 2}$

$ = 0.2165$ N

${B_1} = {{{\mu _0}i} \over {2R}}$

${B_2} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{{3 \over 2}}}}}$

$ \Rightarrow {{{B_1}} \over {{B_2}}} = {1 \over {{R^3}}}{({R^2} + {x^2})^{{3 \over 2}}}$

$ = {1 \over {{R^3}}}(8{R^3})$

$ = 8$

$NiAB = k\theta $

$ \Rightarrow {\theta \over i} = {{NAB} \over k}$

$\Rightarrow$ Sensitivity increases if $B \uparrow $ and $k \downarrow $

$mg \times {1 \over {\sqrt 2 }} = {{ilB} \over {\sqrt 2 }}$

$ \Rightarrow i = {{mg} \over {Bl}}$

$ = {{0.45 \times 10} \over {0.15}} = 30\,A$

$R = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}$

$ \Rightarrow K = {{{B^2}{q^2}{R^2}} \over {2m}}$

$ = {{{{(1.6 \times {{10}^{ - 19}})}^2} \times {{0.6}^2}} \over {2 \times 1.6 \times {{10}^{ - 27}}}}$ J

$= 18$ MeV

${\mu _1} = \pi r_1^2 \times {I_1}$

${\mu _2} = \pi r_2^2 \times {I_2}$

$\therefore$ ${\mu _{net}} = ({\mu _2} - {\mu _1})\left( { - \widehat k} \right)$

$ = \pi \left( {r_2^2 - r_1^2} \right)I\left( { - \widehat k} \right)$

$ = 3.142 \times \left( {{{0.5}^2} - {{0.3}^2}} \right) \times 7\left( { - \widehat k} \right)$

$ = - {7 \over 2}\widehat k$ Am²

As magnetic force is perpendicular to magnetic field

So, $\overrightarrow F $ . $\overrightarrow B $ must be 0

So, 2$\alpha$ $-$ 12 = 0

$\alpha$ = 6

$B = {{{\mu _0}NI} \over {2R}}$

${{{B_X}} \over {{B_Y}}} = {{{N_x}{R_y}} \over {{N_y}{R_x}}}$

$ = {{200 \times 20} \over {400 \times 20}} = {1 \over 2}$

We know that $R = {{mv} \over {Bq}} = \sqrt {{{2mK} \over {Bq}}} $

$\Rightarrow$ Ratio of radii $ = {{{R_1}} \over {{R_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} {{{q_2}} \over {{q_1}}}$

$ \Rightarrow {6 \over 5} = \sqrt {{9 \over 4}} {{{q_2}} \over {{q_1}}}$

$ \Rightarrow {{{q_1}} \over {{q_2}}} = {3 \over 2} \times {5 \over 6} = {5 \over 4}$

In the width of ${r_2} - {r_1}$ total n turns presents.

$\therefore$ In 1 unit width ${n \over {{r_2} - {r_1}}}$ turns presents.

$\therefore$ In the width of dr number of turns,

$n' = {n \over {{r_2} - {r_1}}} \times dr$

Magnetic field (dB) due to element of dr length is

$dB = {{{\mu _0} \times I \times n'} \over {2r}}$

$ = {{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}$

$\therefore$ Total magnetic field due to entire coil is,

$\int {dB = \int_{{r_1}}^{{r_2}} {{{{\mu _0}I} \over {2r}} \times {n \over {({r_2} - {r_1})}}dr} } $

$ \Rightarrow B = {{{\mu _0}In} \over {2({r_2} - {r_1})}}\int_{{r_1}}^{{r_2}} {{{dr} \over r}} $

$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left[ {\log _e^r} \right]_{{r_2}}^{{r_1}}$

$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times \left( {\log _e^{{r_2}} - \log _e^{{r_1}}} \right)$

$ = {{{\mu _0}In} \over {2({r_2} - {r_1})}} \times {\log _e}{{{r_2}} \over {{r_1}}}$

Given,

$f = 10 \times {10^6}$ Hz

$r = 0.6$ m

Charge on proton (q) = e

We know,

Radius $(r) = {{mv} \over {qB}}$

$ = {{\sqrt {2mk} } \over {eB}}$ ..... (1)

Also, we know,

Cyclotron oscillation frequency should be equal to the pendulum evolution frequency.

$\therefore$ $f = {{eB} \over {2\pi m}}$

$ \Rightarrow eB = 2\pi mf$

Putting this value of eB in equation (1) we get

$r = {{\sqrt {2mk} } \over {2\pi mf}}$

$ \Rightarrow {r^2} = {{2mk} \over {4{\pi ^2}{m^2}{f^2}}}$

$ \Rightarrow k = 2{\pi ^2}m{f^2}{r^2}$

$ = 2 \times {\left( {{{22} \over 7}} \right)^2} \times 1.67 \times {10^{ - 27}} \times {\left( {10 \times {{10}^6}} \right)^2} \times {\left( {0.6} \right)^2}$ J

$ = 1.2 \times {10^{ - 12}}$ J

$ = {{1.2 \times {{10}^{ - 12}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV

$ = {{12} \over {16}} \times {10^7}$ eV

$ = 0.75 \times {10^7}$ eV

$ = 7.5 \times {10^6}$ eV

= 7.5 MeV

As B_net $\ne$ 0 that is the wires are carrying current in opposite direction.

${{{\mu _0}I \times 2} \over {2\pi (4 \times {{10}^{ - 2}})}} = 30 \times {10^{ - 6}}$ T

$ \Rightarrow I = {{30 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$ A = 30 A in opposite direction.

Electric field accelerates the particle in the direction of field $\left( {\overrightarrow F = q\overrightarrow E = m\overrightarrow a } \right)$ and magnetic field accelerates the particle perpendicular to the field $\left( {\overrightarrow F = q\overrightarrow v \times \overrightarrow B = m\overrightarrow a } \right)$.

$R = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}$

$ \Rightarrow R \propto \sqrt K $

$\Rightarrow$ ratio = 2 : 1

${{dF} \over {dl}} = 2 \times {10^{ - 6}}$ N/m $ = {{{\mu _0}{i_1}{i_2}} \over {2\pi d}}$

$2 \times {10^{ - 6}} = {{2 \times {{10}^{ - 7}} \times {x^2}} \over {0.2}}$

$x = \sqrt 2 \simeq 1.4$

Inside a hollow cylindrical conductor with uniform current distribution net magnetic field is zero in hollow space.

But outside the cylindrical conductor $B \propto {1 \over r}$

$\Rightarrow$ Graph in option D would be a correct one

Field at P is $ = \left( {{{{\mu _0} \times {i_1}} \over {2\pi {r_1}}} - {{{\mu _0}{i_2}} \over {2\pi {r_2}}}} \right)\left( { - \widehat k} \right)$

$ = - \left( {{{{\mu _0}4} \over {2\pi \times 0.04}} - {{{\mu _0} \times 2} \over {2\pi \times 0.06}}} \right)\widehat k = - {{{\mu _0} \times 200} \over {6\pi }}\widehat k$

So, force $\overrightarrow F = q\overrightarrow v \times \overrightarrow B $

$ = 3\pi \left( {2\widehat i + 3\widehat j} \right) \times \left( { - \left( {{{{\mu _0} \times 200} \over {6\pi }}\widehat k} \right)} \right)$

$ = 3\pi \left( {{{200{\mu _0}} \over {3\pi }}\widehat j - {{100{\mu _0}} \over \pi }\widehat i} \right)$

$ = 200{\mu _0}\widehat j - 300{\mu _0}\widehat i$

$ = 4\pi \times {10^{ - 5}}(2\widehat j - 3\widehat i)$

So, $x = 3$

$B = {\mu _0}ni$

Now $i \to 2i$

And $n \to {n \over 2}$

$B' = {\mu _0}{n \over 2} \times 2i = {\mu _0}ni = B$

$\int {\overline B \,.\,\overline {dl} = {\mu _0}{I_{in}}} $

$ \Rightarrow B \times 2\pi r = {{{\mu _0}I} \over {\pi {R^2}}} \times \pi {r^2}$

$ \Rightarrow B \propto r$

$\therefore$ $r = {{mv} \over {qB}} = {{\sqrt {2m(KE)} } \over {qB}}$

$ \Rightarrow {r_1}:{r_2}:{r_3} = {{\sqrt {{m_1}} } \over {{q_1}}}:{{\sqrt {{m_2}} } \over {{q_2}}}:{{\sqrt {{m_3}} } \over {{q_3}}}$

$ = {{\sqrt 1 } \over 1}:{{\sqrt 2 } \over 1}:{{\sqrt 4 } \over 2}$

$ = 1:\sqrt 2 :1$

Magnetic force $\overrightarrow F \bot \overrightarrow v $

$ \Rightarrow {W_b} = 0$

$ \Rightarrow \Delta KE = 0$ and speed remains constant.

$B = {{{\mu _0}I} \over {2r}}$

${B_a} = {{{\mu _0}I{r^2}} \over {2\left( {{r^2} + {{{r^2}} \over 4}} \right)}}$

$ \Rightarrow {{{B_a}} \over B} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}$

$ \Rightarrow {B_a} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}B$

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

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