Moving Charges and Magnetism

Magnetic field of a circular loop:

$ \begin{aligned} & B_0=\frac{\mu_0 N i}{2 R} \\ & i=\frac{2 R B_0}{\mu_0 N}=\frac{2 \times 5 \times 10^{-2} \times 3.14 \times 10^{-3}}{4 \pi \times 10^{-7} \times 100} \\ & i=2.5 \mathrm{~A} \end{aligned} $

Magnetic moment

$ \begin{aligned} & M=N i A \\ & =100 \times 2.5 \times 3.14 \times\left(5 \times 10^{-2}\right)^2\left[\because A=\pi R^2\right] \\ & =2 \mathrm{~A} \mathrm{~m}^2 \end{aligned} $

For a long straight solid wire carrying steady current, which is uniformly distributed across its crosssection, the variation of magnetic field $(B)$ with distance $(r)$ from axis will be

$ \begin{aligned} & B=\frac{\mu_0 I r}{2 \pi a^2} \Rightarrow B \propto r, \text { for } r < a \\ & B=\frac{\mu_0 I}{2 \pi r} \Rightarrow B \propto \frac{1}{r}, \text { for } r>a \end{aligned} $

$ \begin{aligned} & I_1 R_{A B C}=I_2 R_{A D C} \\ & I_1+I_2=I_0 \\ & I_1 \frac{R}{2}=I_2 R \\ & \frac{I_1}{2}=I_2 \end{aligned} $

$ \begin{aligned} & I_2=\frac{I_0}{3} \\ & I_1=\frac{2 I_0}{3} \end{aligned} $

$ \begin{aligned} & l_1+l_2=l_0 \\ & 2 l_2+l_2=l_0 \end{aligned} $

$ \begin{aligned} & \vec{B}_0=\vec{B}_1+\vec{B}_2 \\ & =\frac{\mu_0 I_1(\cdot)}{4 r}+\frac{\mu_0 I_2(\times)}{4 r} \\ & =\frac{\mu_0 2 I_0}{4 r \times 3}-\frac{\mu_0 I_0}{4 r \times 3}=\frac{\mu_0 I_0}{12 r} \end{aligned} $

Force per unit length on wire $A$ by wire $B$

$ F=\frac{\mu_0}{4 \pi} \frac{2 l \cdot 2 l}{h}=\frac{\mu_0}{4 \pi} \frac{4 l^2}{h} l $

If $F_{n e t} \leq m g$, weight of wire $A$, then wire does not rise from the floor.

So, $F \leq m g$,

$ \begin{aligned} & \frac{\mu_0}{4 \pi} \frac{4 l^2}{h} \leq \lambda g \\ & \therefore h \geq \frac{\mu_0 l^2}{\pi \lambda g} \end{aligned} $

Let the surface charge density be $\sigma=\frac{q}{A}$

Given $\frac{d \sigma}{d t}=$ constant

$\therefore \quad \frac{d}{d t}\left(\frac{q}{A}\right)=\text { constant } \Rightarrow \frac{l}{A} i=\text { constant }$

It means displacement current is constant.

This system will act like a cylindrical wire.

The graph of magnetic field (B) vs $r$ is

To understand the quantized motion of an electron in a uniform magnetic field and determine its magnetic moment in the lowest energy state, consider the following derivations and equations:

Magnetic Force and Centripetal Force:

The magnetic force acting on the electron is counterbalanced by the centripetal force necessary for its circular motion:

$ e v B = \frac{m v^2}{r} $

Solving for the velocity $ v $:

$ v = \frac{e B r}{m} $

Flux through the Electron's Orbit:

The model states that the magnetic flux $ \phi $ through the orbit is linked to Planck’s constant $ h $ as:

$ B \pi r^2 = \frac{n h}{e} $

Rearranging gives:

$ B r^2 = \frac{n h}{e \pi} $

Magnetic Moment:

The magnetic moment $ \mu $ is defined as the current $ I $ times the area $ A $ of the orbit:

$ \mu = I A = \frac{e}{T} \pi r^2 $

Considering current due to orbital motion:

$ \mu = \frac{e v}{2 \pi r} \times \pi r^2 = \frac{e v r}{2} $

Substitute for $ v $ from the earlier velocity equation:

$ \mu = \frac{1}{2} e \left(\frac{e B r}{m}\right) r = \frac{1}{2} e^2 \frac{B r^2}{m} $

Using the flux condition $ B r^2 = \frac{n h}{e \pi} $, we find:

$ \mu = \frac{1}{2} e^2 \frac{n h}{e \pi m} = \frac{n e h}{2 \pi m} $

Lowest Energy State:

For the lowest energy state, take $ n = 1 $:

$ \mu = \frac{e h}{2 \pi m} $

Thus, the magnetic moment of an electron in its lowest energy state is $\frac{e h}{2 \pi m}$.

For no deflection of electron, $\vec{F}_B=\vec{F}_E$

$\begin{aligned} & -e(\vec{v} \times \vec{B})=-e \vec{E} \\ & \Rightarrow \vec{E}=\vec{v} \times \vec{B} \Rightarrow \vec{E} \perp \vec{B} \\ & E=v B=\frac{c}{100} \times 9 \times 10^{-4} \\ & =\frac{3 \times 10^8}{100} \times 9 \times 10^{-4} \\ & =27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1} \end{aligned}$

The magnetic moment of a current-carrying circular loop is given by the formula $ M = I \times A $, where $ I $ is the current and $ A $ is the area of the loop.

Assuming the current ($ I $) is the same for both coils, the magnetic moment is directly proportional to the area of the coil ($ M \propto A $).

For two coils with radii $ r_1 $ and $ r_2 $, the areas are $ A_1 = \pi r_1^2 $ and $ A_2 = \pi r_2^2 $.

Given that the ratio of the radii is $ 1:2 $, we can write:

$ \frac{M_1}{M_2} = \frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $

Thus, the ratio of their respective magnetic moments is $ 1:4 $.

The magnitude of magnetic field due to circular coil of N turns is given by

$ \begin{aligned} & B_C=\frac{\mu_0 i N}{2 R} \\ & =\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1} \\ & =4.4 \times 10^{-3} \mathrm{~T} \\ & =4.4 \mathrm{~mT} \end{aligned} $

According to modified Ampere's law

$\oint B \cdot d I=\mu_0\left(I_C+I_D\right)$

For Loop $L_1 \quad I_C \neq 0$ and $I_D=0$

For Loop $L_2 \quad I_C=0$ and $I_D \neq 0$

Due to $\mathrm{KCL} \quad I_C=I_D$

$\begin{aligned} & \mathrm{m g=i \ell B} \\\\ & 250 \times 10^{-3} \times 9.8=\mathrm{i} \times 2 \times 0.7 \\\\ & \mathrm{i}=\frac{0.250 \times 9.8}{2 \times 0.7}=0.250 \times 7 \\\\ & \mathrm{i}=1.75 \mathrm{~A} \end{aligned} $

Speed of electron will decrease due to electric force magnetic force of electron is zero.

$\vec{F}=I(\vec{\ell} \times \vec{B})$

$ \begin{aligned} & =I[(L \hat{i}) \times(2 \hat{i}+3 \hat{j}-4 \hat{k})] \\\\ & =I(4 L \hat{j}+3 L \hat{k}) \end{aligned} $

$ |\vec{F}|=5 \text { IL } $

$\mathrm{B=\frac{\mu_{0}}{4 \pi} \frac{I}{R}(\pi)-\frac{\mu_{0}}{4 \pi} \frac{2 I}{R}}$

$=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}}\left[1-\frac{2}{\pi}\right]$ outward i.e. away from page.

Magnetic field at the centre of coil

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

On substituting the given values

$ \Rightarrow B = {{4\pi \times {{10}^{ - 7}} \times {{10}^3}} \over {2 \times 62.8 \times {{10}^{ - 2}}}}$

$ = {10^{ - 3}}\,T$

From the right hand curl rule, the shape of magnetic field lines due to long current carrying wire is circular (concentric circles)

Force per unit length between two long current carrying wires $ = {{{\mu _0}{l_1}{l_2}} \over {2\pi d}}$

$F = {{4\pi \times {{10}^{ - 7}} \times 5 \times 10} \over {2\pi \times 10 \times {{10}^{ - 2}}}} = {10^{ - 4}}$ Nm^$-$1

$\Rightarrow$ The force is attractive as direction of current in two conductors is same

Hence, required force is $1 \times {10^{ - 4}}$ Nm^$-$1 and it is attractive.

Magnetic field at the axis of a current carrying circular loop

$|B| = {{{\mu _0}i{a^2}} \over {2{{[{a^2} + {d^2}]}^{3/2}}}} \Rightarrow |B| = {{4\pi \times {{10}^{ - 7}} \times \sqrt 2 \times 1} \over {2{{[1 + 1]}^{3/2}}}}$

$|B| = {{4\pi \times {{10}^{ - 7}} \times \sqrt 2 } \over {2.{{(2)}^{3/2}}}} = {{4\pi \times {{10}^{ - 7}}} \over {{{(2)}^{3/2}} \times {{(2)}^{1/2}}}} = 3.14 \times {10^{ - 7}}$ T

We know, magnetic field at centre of solenoid

$B = {\mu _0}{N \over l}l = {\mu _0}nl$ $\left[ {n = {N \over l}} \right]$

$ = 4\pi \times {10^{ - 7}} \times 100 \times {10^3} \times 1$ $\left[ {n = {{100} \over {{{10}^{ - 3}}}}} \right]$

$ = 4\pi \times {10^{ - 2}}T$

$B = 12.56 \times {10^{ - 2}}T$

According to Biot-Savart's law $d\overrightarrow B = {{{\mu _0}} \over {4\pi }}{{Id\overrightarrow l \times \overrightarrow r } \over {{r^3}}}$ which is applicable for infinitesimal element. It is analogous to Coulomb's law, where $Id\overrightarrow l $ is vector source and electric field is produced by scalar source q. Here statement I is correct and statement II is incorrect.

For solid wire

Inside point

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

$ = {{{\mu _0}Ir} \over {{R^2} \times 2\pi }}$

$B \propto r$

Outside point

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

$B \propto {1 \over r}$