Question

The torque required to hold a small circular coil of $10$ turns, area $1$ $m{m}^2$ 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. $1.5\times {10}^{-6}$N m
• B. $1.5\times {10}^{-8}$N m
• C. $1.5\times {10}^6$N m
• D. $1.5\times {10}^8$N m

Answer 1

Answer: (B)

$1.5\times {10}^{-8}$N m

Here,
For small circular coil,
Number of turns, $N=10$, Area, $A=1m{m}^2=1\times {10}^{-6}{m}^2$
Current, $I_1=\frac{21}{44}A$
For a long solenoid,
Number of turns per metre, $n={10}^3$ per m
Current, $I_2=2.5$A
Magnetic field due to a long solenoid on its axis is
$B={\mu }_{o}n{I}_2$ ..(i)
Magnetic moment of a circular coil is
$M=NA{I}_1$ .(ii)
Torque, $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
$\tau =MB\sin \theta =MB$ ($\because \theta ={90}^o$(Given))
$\tau \left(NA{I}_1\right)\left({\mu }_{o}n{I}_2\right)$ (Using (i) and (ii))
$\tau =10\times 1\times {10}^{-6}\times \frac{21}{44}\times 4\times \frac{22}{7}\times {10}^{-7}\times {10}^3\times 2.5$
$=1.5\times {10}^{-8}$N m