Magnetism

$\begin{aligned} & \mathrm{dM}=\mathrm{dIA} \\ & \mathrm{A}=\pi \mathrm{r}^2=\pi(\mathrm{R} \sin \theta)^2 \\ & \mathrm{dI}=\frac{\mathrm{da}}{\mathrm{T}}=\frac{\sigma(2 \pi \mathrm{r})(\mathrm{R} \mathrm{d} \theta) \omega}{2 \pi} \\ & \mathrm{dI}=\frac{\sigma 2 \pi \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta}{2 \pi}\end{aligned}$

$ \mathrm{dI}=\sigma \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta $

Magnetic dipole moment :

$ \begin{aligned} & M=\int d M=\int_0^\pi \sigma R^2 \omega \pi R^2 \sin ^3 \theta d \theta \\ & M=\sigma R^4 \omega \pi \int_0^\pi \sin ^3 \theta d \theta \quad\left(\because \int_0^\pi \sin ^3 \theta d \theta=\frac{4}{3}\right) \\ & M=\left(\frac{Q}{4 \pi R^2}\right) R^4 \omega \pi\left(\frac{4}{3}\right) \end{aligned} $

Magnetic dipole moment

$ \mathrm{M}=\frac{\mathrm{QR}^2 \omega}{3} $

Angular momentum

$ \begin{aligned} & \mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^2\right) \omega \\ & \frac{\mathrm{M}}{\mathrm{~L}}=\frac{\mathrm{QR}^2 \omega}{3 \times \frac{2}{5} \mathrm{MR}^2 \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right) \end{aligned} $

$\alpha=\frac{5}{3}=1.67$

$ \begin{aligned} &R=0.1 \Omega\\\\ &\varepsilon=\left(\mathrm{B}_1-\mathrm{B}_2\right) b v_{\mathrm{y}}\\\\ &\begin{aligned} & \mathrm{i}=\frac{\varepsilon}{R}=\frac{\mu_0 I}{2 \pi R}\left(\frac{1}{d}-\frac{1}{d+a}\right) b v_y \\\\ & \Rightarrow 10^{-5}=\frac{2 \times 10^{-7} \times 10}{0.1}\left[\frac{1}{4}-\frac{1}{8}\right] \times 2 . v_y \end{aligned} \end{aligned} $

$ \therefore \mathrm{v}_{\mathrm{y}}=2 $


$ \begin{aligned} & \tan \theta=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{1}{\sqrt{3}} \\\\ & \therefore \mathrm{v}_{\mathrm{x}}=2 \sqrt{3} \\\\ & \therefore \mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=4 \end{aligned} $

${r_\alpha } = {{\sqrt {2{m_\alpha }{q_\alpha }V} } \over {{q_\alpha }B}}$

${r_s} = {{\sqrt {2{m_s}{q_s}V} }\over {{q_s}B}}$

${{{r_s}} \over {{r_\alpha }}} = \sqrt {{{{m_s}} \over {{q_s}}}{{{q_\alpha }} \over {{m_\alpha }}}} = \sqrt {\left( {{{32} \over 1}} \right)\left( {{2 \over 4}} \right)} $

${{{r_s}} \over {{r_\alpha }}} = 4$

The magnetic moment of a coil having n turns, area A and current i is given by $\overrightarrow \mu = ni\overrightarrow A $. The torque on this coil when placed in a magnetic field $\overrightarrow B $ is given by $\overrightarrow \tau = \overrightarrow \mu \times \overrightarrow B $. In a moving coil galvanometer, cylindrical magnets ensure that $\overrightarrow A $ and $\overrightarrow B $ are perpendicular to each other. Thus, deflection torque on the coil is $\tau$_d = niAB. The restoring torque on a coil (at deflection angle $\theta$) due to the wire of torsional constant k is $\tau$_r = k$\theta$. In equilibrium, deflection torque is balanced by the restoring torque i.e., $\tau$_d = $\tau$_r, which gives

$i = {k \over {nAB}}\theta $.

The maximum deflection current of the galvanometer (i_g) occurs at full scale deflection $\theta$_max i.e.,

${i_g} = {{k{\theta _{\max }}} \over {nAB}} = {{({{10}^{ - 4}})(0.2)} \over {(50)(2 \times {{10}^{ - 4}})(0.02)}} = 0.01$ A.

A galvanometer of resistance G is converted to an ammeter by connecting a small shunt resistance S in parallel.

Kirchhoff's loop law gives

${i_g}G - (i - {i_g})S = 0$, $ \Rightarrow S = {i_g}G/(i - {i_g})$.

The maximum deflection current of galvanometer sets upper limit on the current measured by this ammeter (i_max). Substitute the values to get

$S = {{{i_g}G} \over {{i_{\max }} - {i_g}}} = {{(0.1)(50)} \over {1 - 0.1}} = {5 \over {0.9}} = 5.55\Omega $.

The charged particle enters the magnetic field region at O. Its velocity is perpendicular to the field direction. The magnetic force on the particle, qv₀B₁, provides centripetal acceleration to move on a circular path from O to P. The radius of the circular path is given by

${r_1} = m{v_0}/(q{B_1})$.

The time taken by the particle to travel from O to P is

${T_1} = \pi {r_1}/{v_0} = \pi m/(q{B_1})$

At P, the particle enters the magnetic field B₂ with its velocity perpendicular to the field direction. The magnetic force qv₀B₂ provides the centripetal acceleration to move in a circular path from P to Q. The radius of the circular path is

${r_2} = m{v_0}/(q{B_2})$.

The time taken by the particle to travel from P to Q is

${T_2} = \pi {r_2}/{v_0} = \pi m/(q{B_2})$.

AT Q, the particle crosses x-axis from below for the first time. Thus, T = T₁ + T₂. The average speed of the particle along the x-axis in the time interval T is given by

$v = {{\int_0^T {{v_x}dt} } \over T} = {{\int_0^{{T_1}} {{v_{x1}}dt + \int_{{T_1}}^{{T_1} + {T_2}} {{v_{x2}}dt} } } \over {{T_1} + {T_2}}}$

$ = {{2{r_1} + 2{r_2}} \over {{T_1} + {T_2}}} = {{{{2m{v_0}} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)} \over {{{\pi m} \over q}\left( {{1 \over {{B_1}}} + {1 \over {{B_2}}}} \right)}}$

$ = {{2{v_0}} \over \pi } = {{2(\pi )} \over \pi } = 2$ m/s

When current in wires is in same direction, the magnetic fields due to two wires all in opposite direction.

From $B = {{{\mu _0}} \over {4\pi }}.{{2I} \over r}$, we get

${B_1} = {{{\mu _0}} \over {4\pi }}.2I\left[ {{1 \over {{x_1}}} - {1 \over {({x_0} - {x_1})}}} \right]$

$ = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} - {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$

$ = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - 2{x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$ ..... (1)

When the direction of current in two wires is opposite, field will be in the same direction.

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{1 \over {{x_1}}} + {1 \over {({x_0} - {x_1})}}} \right]$

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} + {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0}} \over {{x_1}({x_0} - {x_1})}}} \right]$

From ${{m{v^2}} \over r} = qvB$ or $v = {{qBr} \over m}$ or $r = {{mv} \over {qB}}$

$B \propto {1 \over r}$

Therefore, ${{{R_1}} \over {{R_2}}} = {{{B_2}} \over {{B_1}}} = {{{x_0}} \over {({x_0} - 2{x_1})}}$

${{{R_1}} \over {{R_2}}} = {{{x_0}/{x_1}} \over {({x_0}/{x_1}) - (2{x_1}/{x_1})}} = {3 \over {3 - 2}} = 3$

B_R = B_T $-$ B_C

R = Remaining portion

T = Total portion and

C = cavity

${B_R} = {{{\mu _0}{I_T}} \over {2a\pi }} - {{{\mu _0}{I_C}} \over {2(3a/2)\pi }}$ ..... (i)

${I_T} = J(\pi {a^2})$

${I_C} = J\left( {{{\pi {a^2}} \over 4}} \right)$

Substituting the values in Eq. (i), we have

${B_R} = {{{\mu _0}} \over {a\pi }}\left[ {{{{I_T}} \over 2} - {{{I_C}} \over 3}} \right]$

$ = {{{\mu _0}} \over {a\pi }}\left[ {{{\pi {a^2}J} \over 2} - {{\pi {a^2}J} \over {12}}} \right]$

$ = {{5{\mu _0}aJ} \over {12}}$

$\therefore$ $N = 5$

The flux through the ring is

$\phi = B\pi {r^2}$

Assuming the cylinder as a solenoid with close winding, we have

$B = {{{\mu _0}I} \over L}$

Therefore,

$\phi = \left( {{{{\mu _0}I} \over L}} \right)\pi {r^2}\cos 300t$

The induced emf is

$\varepsilon = {{ - d\phi } \over {dt}} = 300\left( {{{{\mu _0}I} \over L}} \right)\pi {r^2}\sin 300t$

Therefore, the current induced is

$i = {\varepsilon \over R} = \left( {{{\pi {r^2}300} \over {RL}}} \right){\mu _0}{I_0}\sin 300t$

The magnetic moment is

M = Current $\times$ Area of loop

Therefore,

$m = \left( {{{{{(3.14)}^2} \times {{(0.1)}^4} \times 300} \over {0.005 \times 10}}} \right){\mu _0}{I_0}\sin 300t$

$ = 6{\mu _0}{I_0}\sin 300t$

Hence, N = 6.

The segments PQ and PR, in the triangle shown here, cannot produce $\overrightarrow B $ at point P since point P lies on them. Here, only QR creates $\overrightarrow B $ at point P.

Applying ${B_P} = {{{\mu _0}i} \over {4\pi R}}(\cos {\phi _1} + \cos {\phi _2})$, where ${\phi _1} = 53^\circ $ and ${\phi _2} = 37^\circ $, we get the magnitude of the magnetic field as follows:

${B_P} = {{{\mu _0}I} \over {4\pi (4\pi \sin 37^\circ )}}(\cos 53^\circ + \cos 37^\circ )$

$ = {{{\mu _0}I} \over {16\pi x(3/5)}}\left( {{{3x} \over {5x}} + {{4x} \over {5x}}} \right)$

$ = {{5{\mu _0}I} \over {48\pi x}}\left( {{7 \over 5}} \right) = {{7{\mu _0}I} \over {48\pi x}} = k\left( {{{{\mu _0}I} \over {48\pi x}}} \right)$

where $k = 7$.