Question

A solenoid of length $0.4m$ and having $500$ turns of wire carries a current of $3amp.$ Calculate the torque required to hold a coil (having radius $0.02cm,$ current $2A$ and turns $50$) in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. $\left({\pi }^2=10\right)$

Answer 1

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

$\overrightarrow{\mu }=IA$

So, $\tau =\left(IA\overrightarrow{B}\right)$

or $\tau =nIAB$

and

$B=\left({\mu }_0{n}_{1}I\right)$
$\tau =\left(nIA\right)\left({\mu }_0{n}_{1}I\right)$

$=\left[50\times 2\times \pi {\left(0.02\times {10}^{-2}\right)}^2\right]\left[500\times 3\times 4\pi \times {10}^7\right]$

$=50\times 2\times {\pi }^2\times {10}^{-4}\times {\left(0.02\right)}^2\times 500\times 3\times 4\times {10}^{-7}$

$\tau =2400\times {10}^{-11}Nm$