# 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. - Physics

Numerical

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 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

#### Solution

Max induced emf = 0.603 V

Average induced emf = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular coil, r = 8 cm = 0.08 m

Area of the coil, A = πr2 = π × (0.08)m2

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3 × 10−2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as:

e = Nω AB

= 20 × 50 × π × (0.08)2 × 3 × 10−2

= 0.603 V

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given as:

"I" = "e"/"R"

= 0.603/10

= 0.0603 A

Average power loss due to joule heating:

"P" = ("eI")/2

= (0.603 xx 0.0603)/2

= 0.018 W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 6 Electromagnetic Induction
Exercise | Q 6.6 | Page 230
NCERT Class 12 Physics Textbook
Chapter 6 Electromagnetic Induction
Exercise | Q 6 | Page 230

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