Q. 94.2( 13 Votes )

# 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° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer :

Given:

Length of side of square, L = 10 cm

Number of turns, n = 20

Current through the square coil, I = 12 A

Angle between the normal to the coil and uniform magnetic field, θ = 30°

Magnitude of magnetic field, B = 0.80 T

The torque experienced by the coil in a magnetic field is given by,

T = n × B × I × A × sin(θ) …(1)

Where,

n = number of turns

B = Strength of magnetic field

I = Current through the coil

A = Area of cross-section of coil

A = L^{2} = 0.1 × 0.1 = 0.01m^{2} …(2)

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

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

T = 20 × 0.80T × 12A × 0.01m^{2} × sin30°

⇒ T = 0.96 Nm

∴ the magnitude torque experienced by the coil is 0.96 N-m.

Hence the torque experienced by the square coil is 0.96 Nm.

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