Question

A solenoid of $0.4m$ length with $500$ turns carries a current of $3A$. A coil of $10$ turns and of radius $0.01m$ carries a current of $0.4A$. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is:

• A. $6{\pi }^2\times {10}^{-7}Nm$
• B. $3{\pi }^2\times {10}^{-7}Nm$
• C. $12{\pi }^2\times {10}^{-7}Nm$
• D. $9{\pi }^2\times {10}^{-7}Nm$

Answer 1

Answer: (A)

Magnetic field at the center of solenoid, $B={\mu }_{o}nI$
Given: $n=\frac{500}{0.4}\frac{turns}{m}I=3A$
$B=4\pi \times {10}^{-7}\times \frac{500}{0.4}\times 3=1.5\pi \times {10}^{-3}T$
Torque on the coil, $\tau =N{I}_{coil}\overrightarrow{A}\times \overrightarrow{B}$
As axis of coil and solenoid are perpendicular to each other, so $\overrightarrow{A}\times \overrightarrow{B}=AB=\pi r^{2}B$
Given: $r=0.01mN=10I_{coil}=0.4A$
$\tau =10\times 0.4\times \pi (0.01{)}^2\times 1.5\pi \times {10}^{-3}$
$⟹\tau =6{\pi }^2\times {10}^{-7}Nm$