Question

The torque required to hold a small circular coil of $10$ turns, are $1m{m}^2$ and carrying a current of $\left[\frac{21}{44}\right]$A in the middle of a long solenoid of ${10}^3$ turns/m carrying a current of $2.5$A, with its axis perpendicular to the axis of the solenoid is:

• A. $1.5\times {10}^{-6}Nm$
• B. $1.5\times {10}^{-8}Nm$
• C. $1.5\times {10}^{6}Nm$
• D. $1.5\times {10}^{8}Nm$

Answer 1

The correct option is B $1.5\times {10}^{-8}Nm$

Given,

No. of turns of coil$=10$

Area$=1m{m}^2$

Current$=\frac{21}{44}A$

$\therefore$ Dipole moment $\mu =NAI=10\times {10}^{-6}\times \frac{21}{44}A{m}^2$

$=4.7\times {10}^{-6}A{m}^2$

Given, No. of turns per $m={10}^3\frac{turns}{m},I=2.5A$

Magnetic field due to a solenoid $\overrightarrow{B}={\mu }_{0}NI=3.14\times {10}^{-3}T$

$\therefore Torque=\mu B\sin \theta =1.499\times {10}^{-8}Nm\times \sin {90}^{\circ }$

$=1.5\times {10}^{-8}Nm$