Moving Charges and Magnetism

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} $