Question

A circular coil of radius 8.0 cm and 20 turns is rotated about itsvertical diameter with an angular speed of 50 rad s–1 in a uniformhorizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain themaximum and average emf induced in the coil. If the coil forms aclosed loop of resistance 10 W, calculate the maximum value of currentin the coil. Calculate the average power loss due to Joule heating.Where does this power come from?

Answer 1

Given: The radius of the circular coil is 8 cm, the numbers of turns are 20, the angular speed is 50 rad/s, the uniform magnetic field is 3× 10 −2  T and the resistance of the loop is 10 Ω.

The area of the coil is given as,

A=π r 2

Where, the radius of the circular coil is r.

By substituting the given values in the above formula, we get

A=π× ( 0.08 ) 2

The maximum induced emf is given as,

e=BANω

Where, the magnetic field strength is B, the area of the coil is A, the number of turns are N and the angular frequency is ω.

By substituting the given values in the above formula, we get

e=3× 10 −2 × ( 0.08 ) 2 ×20×50 =0.603 V

Thus, the maximum induced emf is 0.603 V.

The average emf induced in the coil over a full cycle will be zero.

The maximum current in the coil is given as,

I= e R

Where, the resistance is R.

By substituting the given values in the above formula, we get

I= 0.603 10 =0.0603 A

Thus, the maximum current in the coil is 0.0603 A.

The average power loss due to joule heating is given as,

P= eI 2

Where, the average power loss is P.

By substituting the given values in the above formula, we get

P= 0.603×0.0603 2 =0.018 W

Thus, the average power loss is 0.018 W and the induced current opposes the rotation of the coil so, an external rotor must supply the extra torque to counter this torque and keep the rotation constant. Thus, the external rotor is the source of this dissipated power.