Question

A solenoid of $0.4\text{m}$ length with $500$ turns carries a current of $3\text{A}.$ A coil of $10$ turns and of radius $0.01\text{m}$ carries of $0.4\text{A}$. The torque required to hold the coil with its axis at right angle to that of the solenoid is :

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

Answer 1

The correct option is A $6{\pi }^2\times {10}^{-7}\text{Nm}$

Given:

Length of the solenoid $\left(l\right)=0.4\text{m}$

number of turns $\left(N\right)=500$

Current $\left(I\right)=3\text{A}$

The magnetic field at the centre of the solenoid will be

$B={\mu }_{0}nI$ (Since, solenoid is long enough compared to the coil taken in the question)

$\Rightarrow B={\mu }_0\left(\frac{N}{l}\right)I[\because n=\frac{N}{l}]$

$\Rightarrow B=4\pi \times {10}^{-7}\times \frac{500}{0.4}\times 3$

$\Rightarrow B=15\pi \times {10}^{-4}\text{T}$

Torque acting on a current carrying loop is given by,
$$\overrightarrow{\tau }=NI(\overrightarrow{A}\times \overrightarrow{B})$$$\tau =\left(NIA\right)B\sin \theta$

Given,

$N=10,i=0.4\text{A}$

As axis of coil and solenoid are perpendicular to each other,
$\therefore \theta ={90}^{\circ }$

$A=\pi r^2=\pi \times (0.01{)}^2=\pi \times {10}^{-4}{\text{m}}^2$

$\Rightarrow \tau =(10\times 0.4\times \pi \times {10}^{-4})\times 15\pi \times {10}^{-4}\times 1$

$\Rightarrow \tau =6{\pi }^2\times {10}^{-7}\text{Nm}$

So, the torque required to hold the coil will be equal in magnitude and opposite in direction to the torque acting on the coil due to the magnetic field produced by the solenoid.

Hence, option (a) is the correct answer.