# 9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of  $30 \degree$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by

$\tau =nBIAsin\theta$

where n = number of turns, I is the current in the coil, A is the area of the coil and $\theta$ is the angle between the magnetic field and the vector normal to the plane of the coil.

In the given question n = 20, B=0.8 T, A=0.1$\times$0.1=0.01 m2, I=12 A, $\theta$=30o

$\tau =20\times 0.8\times 12\times 0.01\times sin30^{o}$

=0.96 Nm

The coil therefore experiences a torque of magnitude 0.96 Nm.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-