Moving Charges and Magnetism — NCERT Solutions

For a long solenoid,

$B=\mu_0 n I$

Total turns $N=5\times400=2000$.
Length $l=80\,\text{cm}=0.80\,\text{m}$.

So,
$$n=\frac{N}{l}=\frac{2000}{0.80}=2500\,\text{turns m}^{-1}$$

Given $I=8.0\,\text{A}$.

$$B=4\pi\times10^{-7}\times2500\times8$$

$$=8\pi\times10^{-3}\approx2.5\times10^{-2}\,\text{T}$$

So the magnetic field is about $2.5\times10^{-2}\,\text{T}$.

Torque on a current loop is

$$\tau = N I A B \sin\theta$$

Here:

• $N=20$
• $I=12\,\text{A}$
• side $=10\,\text{cm}=0.10\,\text{m}$, so area $A=(0.10)^2=0.01\,\text{m}^2$
• $B=0.80\,\text{T}$
• angle between normal and field $\theta=30^\circ$

$$\tau=20\times12\times0.01\times0.80\times\sin30^\circ$$

$$=20\times12\times0.01\times0.80\times\frac12 =0.96\,\text{N m}$$

So the torque experienced by the coil is $0.96\,\text{N m}$.

For a moving coil meter, the current sensitivity is

$$\frac{\phi}{I}=\frac{NAB}{k}$$

Since the spring constants are identical, compare the products $NAB$.

For $M_1$:
$$N_1A_1B_1=30\times 3.6\times10^{-3}\times 0.25$$
$$=30\times 0.9\times10^{-3}=27\times10^{-3}$$

For $M_2$:
$$N_2A_2B_2=42\times 1.8\times10^{-3}\times 0.50$$
$$=42\times 0.9\times10^{-3}=37.8\times10^{-3}$$

So,
$$\frac{(\phi/I)_2}{(\phi/I)_1}=\frac{37.8}{27}=1.4=\frac{7}{5}$$

The ratio is $7:5$.

For the voltage sensitivity,
$$\frac{\phi}{V}=\left(\frac{NAB}{k}\right)\frac{1}{R}$$

So compare $\dfrac{NAB}{R}$.

For $M_1$:
$$\frac{N_1A_1B_1}{R_1}=\frac{27\times10^{-3}}{10}=2.7\times10^{-3}$$

For $M_2$:
$$\frac{N_2A_2B_2}{R_2}=\frac{37.8\times10^{-3}}{14}=2.7\times10^{-3}$$

Thus,
$$\frac{(\phi/V)_2}{(\phi/V)_1}=1$$

So the voltage sensitivity ratio is $1:1$.

The electron moves in a circle because the magnetic force is always perpendicular to its velocity. Hence it changes only the direction of motion, not the speed, and acts as the centripetal force.

For circular motion,

$$r=\frac{mv}{qB}$$

Given:

• $B=6.5\,\text{G}=6.5\times10^{-4}\,\text{T}$
• $v=4.8\times10^{6}\,\text{m s}^{-1}$
• $q=1.5\times10^{-19}\,\text{C}$
• $m=9.1\times10^{-31}\,\text{kg}$

$$r=\frac{9.1\times10^{-31}\times4.8\times10^{6}}{1.5\times10^{-19}\times6.5\times10^{-4}}$$

$$=\frac{4.368\times10^{-24}}{9.75\times10^{-23}} \approx 4.48\times10^{-2}\,\text{m}$$

So the radius is about $4.5\times10^{-2}\,\text{m}$, i.e. $4.5\,\text{cm}$.

The frequency of revolution is

$$\nu=\frac{qB}{2\pi m}$$

Given:

• $q=1.5\times10^{-19}\,\text{C}$
• $B=6.5\times10^{-4}\,\text{T}$
• $m=9.1\times10^{-31}\,\text{kg}$

$$\nu=\frac{1.5\times10^{-19}\times6.5\times10^{-4}}{2\pi\times9.1\times10^{-31}}$$

$$\approx \frac{9.75\times10^{-23}}{5.72\times10^{-30}} \approx 1.7\times10^{7}\,\text{Hz}$$

So the frequency is about $1.7\times10^7\,\text{Hz}$.

It does not depend on the speed of the electron, because $\nu=\dfrac{qB}{2\pi m}$ contains only $q$, $B$, and $m$.

Torque on a current loop is

$$\tau = N I A B \sin\theta$$

Given:

• $N=30$
• $I=6.0\,\text{A}$
• $r=8.0\,\text{cm}=0.08\,\text{m}$
• $A=\pi r^2=\pi(0.08)^2=0.0201\,\text{m}^2$
• $B=1.0\,\text{T}$
• $\theta=60^\circ$

$$\tau=30\times6.0\times0.0201\times1.0\times\sin60^\circ$$

$$=180\times0.0201\times0.866 \approx 3.13\,\text{N m}$$

So the counter torque required is about $3.1\,\text{N m}$.

No, the answer would not change. For a planar current loop, the torque depends on the area enclosed, not on the exact shape, as long as the area is the same and other conditions remain unchanged. In a given perimeter, a circle encloses the greatest area, so a flexible loop tends to become circular to maximise flux.