Question

A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 amp. If the torque required to hold a coil (having radius 0.02 cm, current 2A and turns 50) in the middle of the solenoid with its axis perpendicular to the axis of the solenoid $\frac{3}{x} \times 10^{-7}$ Nm, find x. (Take $\pi^2 = 10$)

• A. 5
• B. 10
• C. 2
• D. 1

Answer 1

Answer: (A)

5

Answer 2

The magnetic field ($B$) inside a long solenoid is given by $B = \mu_0 n I_s = \mu_0 \frac{N_s}{L} I_s$. The magnetic dipole moment ($M$) of the coil is given by $M = N_c I_c A_c = N_c I_c \pi r_c^2$. The torque ($\tau$) on a magnetic dipole in a magnetic field is $\tau = MB \sin\theta$. Given that the coil's axis is perpendicular to the solenoid's axis, $\theta = 90^\circ$, so $\sin\theta = 1$, and $\tau = MB$.

Given values: Solenoid: $L = 0.4$ m, $N_s = 500$, $I_s = 3$ A. $\mu_0 = 4\pi \times 10^{-7}$ Tm/A. Coil: $r_c = 0.02$ cm $= 2 \times 10^{-4}$ m, $N_c = 50$, $I_c = 2$ A. Torque: $\tau = \frac{3}{x} \times 10^{-7}$ Nm. Approximation: $\pi^2 = 10$.

1. Magnetic field ($B$) of the solenoid: $B = (4\pi \times 10^{-7}) \times \frac{500}{0.4} \times 3 = 4\pi \times 10^{-7} \times 1250 \times 3 = 15000\pi \times 10^{-7} = 1.5\pi \times 10^{-3}$ T.

2. Magnetic dipole moment ($M$) of the coil: Area of the coil $A_c = \pi r_c^2 = \pi (2 \times 10^{-4})^2 = 4\pi \times 10^{-8}$ m$^2$. $M = 50 \times 2 \times (4\pi \times 10^{-8}) = 400\pi \times 10^{-8} = 4\pi \times 10^{-6}$ Am$^2$.

3. Torque ($\tau$) on the coil: $\tau = MB = (4\pi \times 10^{-6}) \times (1.5\pi \times 10^{-3}) = 6\pi^2 \times 10^{-9}$ Nm. Using $\pi^2 = 10$: $\tau = 6 \times 10 \times 10^{-9} = 60 \times 10^{-9} = 6 \times 10^{-8}$ Nm.

4. Solve for x: $6 \times 10^{-8} = \frac{3}{x} \times 10^{-7}$ $x = \frac{3 \times 10^{-7}}{6 \times 10^{-8}} = \frac{3}{6} \times 10^1 = 0.5 \times 10 = 5$.