Answer :


Number of turns in the coil, n = 30

Radius of coil, r = 8 cm

Current through the coil, I = 6.0 A

Strength of magnetic field = 1.0 T

Angle between the direction of field and normal to coil, θ = 60°

We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.

Torque on the coil due to magnetic field is given by,

T = n × B × I × A × sinθ …(1)


n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = πr2 = 3.14 × (0.08 × 0.08) = 0.0201m2 …(2)

θ = Angle between normal to cross-section of coil and magnetic field

Now, by putting the values in equation (1) we get,

T = 30 × 6.0T × 1A × 0.0201m2 × sin60°

T = 3.133 Nm

Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

A bar magnet of mPhysics - Board Papers

Consider a circulPhysics - Exemplar

(a) Derive the exPhysics - Board Papers

A circular currenPhysics - Exemplar

A rectangular conPhysics - Exemplar

A uniform conductPhysics - Exemplar

Would your answerNCERT - Physics Part-I

A uniform magnetiNCERT - Physics Part-I

A circular coil oNCERT - Physics Part-I