Q. 63.9( 32 Votes )

# A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^{–1} in a uniform horizontal magnetic field of magnitude 3.0 × 10^{–2} T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 W, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating.

Where does this power come from?

Answer :

Given:

Radius of the circular coil r = 8.0 cm = 0.08 m

No. of turns in the coil, N =20

Angular frequency ω = 50 rad s ^{- 1}

Magnitude of magnetic field B =3.0 × 10^{–2} T

Resistance of the closed loop = 10 Ω

The area of the coil can be calculated using the formula

A =πr^{2}

A = 3.14 × (0.08 m)^{2}

The maximum induced e.m.f is calculated as follows:

e = N ω A B

e = 20 × 50 rad s ^{- 1} × 3.14 × (0.08 m)^{2} × 3.0 × 10^{–2} T

On calculating we get

e = 0.603 V

Since for a full cycle, the average induced emf will be zero.

The maximum current for the circular coil can be calculated as follows:

I = e/R

Substituting the values

I = 0.603 V/10 Ω

I = 0.0603A

The average power loss due to Joule’s heating effect in the circular coil will be given by:

P = (eI)/2

P = (0.603 V × 0.0603 A)/2

On calculating we get

P = 0.018 W

A torque was produced due to current induced in the coil. This torque opposes the rotation of the circular coil. Since a rotor is the external agent in the coil and it should apply the torque so that the coil keeps rotating in the uniform manner. Therefore, the dissipated power comes from the external rotor.

Rate this question :