Q. 13 A4.0( 22 Votes )

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Answer :

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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