Moving Charges and Magnetism — NCERT Solutions

Given:

• Length of solenoid, $L = 80\,\text{cm} = 0.80\,\text{m}$
• Number of layers = 5
• Turns per layer = 400
• Total number of turns, $N = 5 \times 400 = 2000$
• Current, $I = 8.0\,\text{A}$
• $\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}$

Number of turns per unit length:
$$n = \frac{N}{L} = \frac{2000}{0.80} = 2500\,\text{turns/m}$$

Formula used:
The magnetic field inside a solenoid near its centre:
$$B = \mu_0 n I$$

Calculation:
$$B = 4\pi \times 10^{-7} \times 2500 \times 8.0$$
$$B = 4\pi \times 10^{-7} \times 20000$$
$$B = 4\pi \times 2 \times 10^{-3}$$
$$B = 8\pi \times 10^{-3}$$
$$B \approx 2.512 \times 10^{-2}\,\text{T}$$
$$B \approx 0.025\,\text{T}$$

(Note: The diameter of the solenoid is not needed for this calculation as the field inside an ideal solenoid does not depend on its cross-sectional dimensions.)

Answer: The magnitude of the magnetic field inside the solenoid near its centre is $B \approx 2.5 \times 10^{-2}\,\text{T}$.

Given:

• Side of square coil, $a = 10\,\text{cm} = 0.10\,\text{m}$
• Number of turns, $N = 20$
• Current, $I = 12\,\text{A}$
• Angle between normal to coil and magnetic field, $\theta = 30°$
• Magnetic field, $B = 0.80\,\text{T}$

Area of the coil:
$$A = a^2 = (0.10)^2 = 0.01\,\text{m}^2$$

Magnetic moment:
$$m = NIA = 20 \times 12 \times 0.01 = 2.4\,\text{A m}^2$$

Formula used:
The torque on a current-carrying coil in a magnetic field:
$$\tau = NIA B\sin\theta = mB\sin\theta$$

Calculation:
$$\tau = 2.4 \times 0.80 \times \sin 30°$$
$$\tau = 2.4 \times 0.80 \times 0.5$$
$$\tau = 0.96\,\text{N m}$$

Answer: The magnitude of the torque experienced by the coil is $\tau = 0.96\,\text{N m}$.

Given:

• $R_1 = 10\,\Omega,\; N_1 = 30,\; A_1 = 3.6 \times 10^{-3}\,\text{m}^2,\; B_1 = 0.25\,\text{T}$
• $R_2 = 14\,\Omega,\; N_2 = 42,\; A_2 = 1.8 \times 10^{-3}\,\text{m}^2,\; B_2 = 0.50\,\text{T}$
• Spring constants: $k_1 = k_2 = k$ (identical)

Concept:
For a moving coil galvanometer, the deflection is given by $k\phi = NIAB$.

• Current sensitivity $= \dfrac{\phi}{I} = \dfrac{NAB}{k}$
• Voltage sensitivity $= \dfrac{\phi}{V} = \dfrac{NAB}{kR}$

(a) Ratio of current sensitivity:
$$\frac{(I_s)_2}{(I_s)_1} = \frac{N_2 A_2 B_2 / k}{N_1 A_1 B_1 / k} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1}$$
$$= \frac{42 \times 1.8 \times 10^{-3} \times 0.50}{30 \times 3.6 \times 10^{-3} \times 0.25}$$
$$= \frac{42 \times 1.8 \times 0.50}{30 \times 3.6 \times 0.25}$$
$$= \frac{37.8}{27.0} = 1.4$$

$$\boxed{\frac{(I_s)_2}{(I_s)_1} = 1.4}$$

(b) Ratio of voltage sensitivity:
$$\frac{(V_s)_2}{(V_s)_1} = \frac{N_2 A_2 B_2 / (k R_2)}{N_1 A_1 B_1 / (k R_1)} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1} \times \frac{R_1}{R_2}$$
$$= 1.4 \times \frac{10}{14} = 1.4 \times \frac{5}{7} = 1.0$$

$$\boxed{\frac{(V_s)_2}{(V_s)_1} = 1.0}$$

Answer:
(a) The ratio of current sensitivity of M₂ to M₁ is 1.4.
(b) The ratio of voltage sensitivity of M₂ to M₁ is 1.0 (they are equal).

Given:

• Magnetic field, $B = 6.5\,\text{G} = 6.5 \times 10^{-4}\,\text{T}$
• Speed of electron, $v = 4.8 \times 10^6\,\text{m s}^{-1}$
• Charge of electron, $e = 1.6 \times 10^{-19}\,\text{C}$
• Mass of electron, $m_e = 9.1 \times 10^{-31}\,\text{kg}$
• The velocity is normal to the field.

Why the path is circular:
The magnetic force on the electron is $F = evB\sin\theta$. Since $v \perp B$, $\theta = 90°$, so $F = evB$. This force is always perpendicular to the velocity of the electron (it is a centripetal force). Since the force does no work (it is always perpendicular to displacement), the speed remains constant. The direction of force changes continuously, always pointing towards the centre of the circular path. Hence, the electron moves in a circular orbit.

Radius of circular orbit:
The magnetic force provides the centripetal force:
$$evB = \frac{m_e v^2}{r}$$
$$r = \frac{m_e v}{eB}$$

Calculation:
$$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$$
$$r = \frac{9.1 \times 4.8 \times 10^{-25}}{1.6 \times 6.5 \times 10^{-23}}$$
$$r = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}}$$
$$r = \frac{43.68}{10.4} \times 10^{-2}$$
$$r = 4.2 \times 10^{-2}\,\text{m} = 4.2\,\text{cm}$$

Answer: The path of the electron is circular because the magnetic force is always perpendicular to its velocity. The radius of the circular orbit is $r \approx 4.2\,\text{cm}$.

Given (from Exercise 4.11):

• $B = 6.5 \times 10^{-4}\,\text{T}$
• $e = 1.6 \times 10^{-19}\,\text{C}$
• $m_e = 9.1 \times 10^{-31}\,\text{kg}$

Formula used:
The cyclotron frequency (frequency of revolution) is:
$$\nu_c = \frac{eB}{2\pi m_e}$$

Calculation:
$$\nu_c = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}}$$
$$\nu_c = \frac{1.04 \times 10^{-22}}{5.7196 \times 10^{-30}}$$
$$\nu_c = \frac{1.04}{5.7196} \times 10^{8}$$
$$\nu_c \approx 1.82 \times 10^{7}\,\text{Hz} \approx 18\,\text{MHz}$$

Does the frequency depend on speed?
No, the frequency does not depend on the speed of the electron. This is because:

• The radius $r = \dfrac{m_e v}{eB}$ is proportional to $v$.
• The time period $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m_e}{eB}$, which is independent of $v$.
• Therefore, $\nu_c = \dfrac{1}{T} = \dfrac{eB}{2\pi m_e}$ is independent of the speed $v$.

This is the fundamental principle exploited in the cyclotron.

Answer: The frequency of revolution of the electron is $\nu_c \approx 1.82 \times 10^7\,\text{Hz}$. The frequency does not depend on the speed of the electron.

Part (a):

Given:

• Number of turns, $N = 30$
• Radius of coil, $R = 8.0\,\text{cm} = 0.08\,\text{m}$
• Current, $I = 6.0\,\text{A}$
• Magnetic field, $B = 1.0\,\text{T}$
• Angle between field lines and normal to coil, $\theta = 60°$

Area of the coil:
$$A = \pi R^2 = \pi \times (0.08)^2 = \pi \times 6.4 \times 10^{-3} = 2.011 \times 10^{-2}\,\text{m}^2$$

Torque on the coil:
$$\tau = NIAB\sin\theta$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times 1.0 \times \sin 60°$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times \frac{\sqrt{3}}{2}$$
$$\tau = 30 \times 6.0 \times 2.011 \times 10^{-2} \times 0.866$$
$$\tau = 180 \times 2.011 \times 10^{-2} \times 0.866$$
$$\tau = 180 \times 1.7415 \times 10^{-2}$$
$$\tau \approx 3.13\,\text{N m}$$

To prevent the coil from turning, the counter torque must be equal in magnitude:
$$\tau_{\text{counter}} \approx 3.13\,\text{N m}$$

Part (b):

No, the answer would not change. The torque on a planar current loop depends only on the magnetic moment $\mathbf{m} = NIA$, the magnetic field $B$, and the angle $\theta$ between $\mathbf{m}$ and $\mathbf{B}$. It does not depend on the shape of the loop, only on the area enclosed. Since the irregular planar coil encloses the same area $A$, carries the same current $I$, has the same number of turns $N$, and is placed in the same field at the same angle, the torque (and hence the required counter torque) remains the same at $\approx 3.13\,\text{N m}$.

Answer:
(a) The magnitude of the counter torque required is $\tau \approx 3.13\,\text{N m}$.
(b) No, the answer does not change, because torque depends only on the area of the loop, not its shape.