Question

The torque required to hold a small circular coil of 10 turns, area $1m{m}^2$ and carrying a current of $\frac{21}{44}A$ at 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 solenoid is

• A. $1.5\times {10}^9$Nm
• B. $1.5\times {10}^{-9}$Nm
• C. $1.5\times {10}^{-8}$Nm
• D. $1.5\times {10}^{-6}$nm

Answer 1

Answer: (C)

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

$\text{Given}:$ For small circular coil,Number of turns,N = 10, Area A=1 $m{m}^2=1\times {10}^{-6}{m}^2$, Current, $I_1=\frac{21}{44}A$

For a long solenoid,Number of turns per metre, $n={10}^3/m$, Current, $I_2=2.5A$

$\text{Solution}:$

Magnetic field due to a long solenoid on its axis is

$B={\mu }_{0}n{I}_2$ ...(i)

Magnetic moment of a circular coil is

$M=NA{I}_1$...(ii)

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

$\tau =MBsin\theta =MB(because\theta ={90}^0)$

$\tau =\left(NA{I}_1\right)\left({\mu }_{0}n{I}_2\right)$ (Using(i)and (ii))

$\tau =10\times 1\times {10}^{-6}\times \frac{21}{44}\times 4\times \frac{22}{7}\times {10}^{-7}\times {10}^3\times 2.5=1.5\times {10}^{-8}Nm$

$\text{Hence C is the correct option}$