Q. 13 A4.0( 25 Votes )

# A circular coil o

Given:

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)

Where,

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.

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