Moving Charges and Magnetism

Given, radius of hollow cylinder = R

Current = i

Distance from axis = r

Let magnetic field be B.

$\because$ No current is passing through region r < R

and magnetic field $(B)=\frac{1}{2\pi}\frac{\propto_0I}{r}$

$\begin{aligned} \text{So,} \\& B(r < R)=0 \\ & B(r=R)=\frac{1}{2 \pi} \frac{\propto_0 I}{R} \\ \text {and}\\& B(r>R)=\frac{1}{2 \pi} \frac{\propto_0 I}{r} \end{aligned}$

So, these variations are correctly shown as below.

Force on charge particle in magnetic field is balanced by centripetal force, so ${{m{v^2}} \over r} = qvB\sin 90^\circ $

[$\theta$ = 90$^\circ$, therefore magnetic field is perpendicular]

${{m{v^2}} \over r} = qvB$ [sin90$^\circ$ = 1]

$ \Rightarrow r = {{mv} \over {qB}}$

For both particles, q and B are same.

Therefore, r $\propto$ mv

or, ${{{r_A}} \over {{r_B}}} = {{{m_A}{v_A}} \over {{m_B}{v_B}}}$

As, r_A > r_B [given]

$ \Rightarrow {m_A}{v_A} > {m_B}{v_B}$

Given, mass of the wire,

$ m=0.2 \mathrm{~kg} $

Length of the wire, $l=1.5 \mathrm{~m}$

Current, $I=2 \mathrm{~A}$

The force acting on the current carrying wire in uniform magnetic field,

$ \begin{aligned} & F=I B l \sin \theta \\ & F=I B l \sin 90^{\circ} \quad\left(\because \theta=90^{\circ}\right) \end{aligned} $

$ \begin{equation*} F=I B l \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Weight of the wire, $w=m g=0.2 \times 10 \mathrm{~N}$ In the position of suspension,

$ B I l=m g \Rightarrow B=\frac{m g}{I l}=\frac{0.2 \times 10}{2 \times 1.5}=0.67 \mathrm{~T} $

The magnetic field at some internal point on the axis of a solenoid is given by

$ B=\frac{\mu_0 n I}{2}\left(\cos \theta_1-\cos \theta_2\right) $

Magnetic field at the centre of a long solenoid is given by setting $\theta_1=0^{\circ}$ and $\theta_2=\pi$.

$ \begin{align*} \Rightarrow \quad B_1 & =\frac{\mu_0 n I}{2}\left(\cos 0^{\circ}-\cos \pi\right) \\ & =\frac{\mu_0 n I}{2}[1-(-1)] \\ & =\mu_0 n I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{align*} $

Similarly, the magnetic field at the end of a long solenoid is given by setting $\theta_1=\frac{\pi}{2}$ and $\theta_2=\pi$.

$ \begin{align*} \Rightarrow \quad B_2 & =\frac{\mu_0 n I}{2}\left(\cos \frac{\pi}{2}-\cos \pi\right) \\ & =\frac{\mu_0 n I}{2}[0-(-1)] \\ & =\frac{\mu_0 n I}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eq. (i) and Eq. (ii), we get

$ \frac{B_1}{B_2}=\frac{\mu_0 n I}{\mu_0 n I} \times 2=2 $

The magnetic field at point due to a current carrying straight conductor of infinite length is

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

The magnetic flux through circular loop around conductor in $x y$-plane is

$ \phi_B=B A=\frac{\mu_0 I}{2 \pi R} \times \pi R^2=\frac{\mu_0 I R}{2} $

The given situation is shown in the following figure

Magnetic field at point $P$ due to the wire having current $I$ is given as

$ B_I=\frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y} $   (downward)

Magnetic field at point $P$ due to the wire having current $i$ is given as

$ B_i=\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} $     (upward)

According to question, net magnetic field at point $P$ is zero, hence

$ \begin{array}{rlrl} & B_I =B_i \\ & \Rightarrow \frac{\mu_0}{2 \pi} \cdot \frac{I}{x+y} =\frac{\mu_0}{2 \pi} \cdot \frac{i}{y} \\ & \Rightarrow \frac{I}{x+y} =\frac{i}{y} \Rightarrow i=I\left(\frac{y}{x+y}\right) \end{array} $

For a solenoid,

$ I=20 \mathrm{~A}, l=2 \mathrm{~m}, B=20 \mathrm{mT}=20 \times 10^{-3} \mathrm{~T} $

Diameter, $d=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$

∴ Radius, $r=\frac{3}{2} \times 10^{-2} \mathrm{~m}$

If $n$ be the number of turns per unit length in the solenoid, then

$ \begin{aligned} B & =\mu_0 n I \Rightarrow n=\frac{B}{\mu_0 I} \\ & =\frac{20 \times 10^{-3}}{4 \pi \times 10^{-7} \times 20}=\frac{10^4}{4 \pi} \end{aligned} $

∴ Total number of turns in the solenoid,

$ N=n l=\frac{10^4}{4 \pi} \times 2 \Rightarrow N=\frac{10^4}{2 \pi} $

Length of wire $=N \times$ length of wire in one turn

$ \begin{aligned} & =N \times 2 \pi r=\frac{10^4}{2 \pi} \times 2 \pi \times \frac{3}{2} \times 10^{-2} \\ & =1.5 \times 10^2 \mathrm{~m}=150 \mathrm{~m} \end{aligned} $

The given situation is shwon below

Current through tangent galvanometer and deflection given by it are related as,

$ I=\frac{2 r}{\mu_0 N} \cdot B_H \cdot \tan \theta $

where, $r=$ radius of coil of galvanometer.

If $R=$ resistance of galvanometer, then potential drop across it is

$ V=I R=\frac{2 r \cdot B_H \cdot \tan \theta \cdot R}{\mu_0 \cdot N} $

In given case, $V_A=V_B$

$ \begin{gathered} \Rightarrow \frac{2 r_1 \tan \theta_1}{N_1}=\frac{2 r_2 \tan \theta_2}{N_2} \\ \quad\left[\because R_1=R_2=R=10 \Omega\right] \\ N_2=\frac{N_1 \cdot r_2 \cdot \tan \theta_2}{r_1 \cdot \tan \theta_1} \\ \quad=\frac{2 \times 16 \times 10^{-2} \times \tan 60^{\circ}}{8 \times 10^{-2} \times \tan 30^{\circ}} \\ \quad=2 \times 2 \times 3=12 \text { turns } \end{gathered} $

Ratio of magnitudes of magnetic moment and angular momentum of rotating charged particle in an orbit is called its 'Gyromagnetic ratio' its' value is equals to $1 / 2$ of specific charge of the particle.

$ \text { ∴ Gyromagnetic ratio }=\frac{M}{L}=\frac{1}{2}\left(\frac{q}{m}\right) $

Magnitude of torque on the coil is given as

$ \tau=|\mathbf{m} \times \mathbf{B}| \Rightarrow \tau=N I A B \sin \theta $

where, $\theta$ is the angle between axis of loop (normal to the loop) and direction of magnetic field.

Here, $\theta=0 \quad$ So, $\quad \tau=0$

For a solenoid,

$ B=\mu_0 n I \Rightarrow n=\frac{N}{L} $

Here, $B=4 \pi \times 10^{-3} \mathrm{~T}$

$ n=\frac{400}{50 \times 10^{-2}}=\frac{400 \times 100}{50}=800 \mathrm{turns} / \mathrm{m} $

$ \begin{aligned} & \text { Since, } B=\mu_0 n I \\ & 4 \pi \times 10^{-3}=4 \pi \times 10^{-7} \times 800 \times I \\ & \Rightarrow \quad I=\frac{4 \pi \times 10^{-3}}{4 \pi \times 10^{-7} \times 800} \\ & \Rightarrow \quad I=\frac{10000}{800}=12.5 \mathrm{~A} \end{aligned} $

Given, current in the wire, I = 10A

Radius of the wire, R = 20$\pi$ cm = 20$\pi$ $\times$ 10^$-$2 m

Magnetic field produced by semi-circular current carrying thin wire at centre O,

$B = {{{\mu _0}I} \over {4R}}$

Substituting all the given values in above equation, we get

$B = {{4\pi \times {{10}^{ - 7}} \times 10} \over {4 \times 20\pi \times {{10}^{ - 2}}}}$

B = 5 $\times$ 10^$-$6 T

Now, magnitude of force per unit length,

F = IB = 10 $\times$ 5 $\times$ 10^$-$6

= 50 $\times$ 10^$-$6 N/m

= 50 $\mu$N/m