Test Index

NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

© examsnet.com
Question : 6
Total: 17
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 ohm, 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:  
If the circular coil rotates in the magnetic field B at an angular velocity ω, then instantaneous induced emf can be calculated.

Instantaneous flux φ = BA cosωt
emf , ε = -
dϕ
dt
 = - BA
dcos(ωt)
dt
e = – NBA [– ω sinωt] or e = NBAω sin ωt
Max. emf for sin ωt = 1
εmax = NBAω = 3 × 10–2 × π(8×10–2)2 × 50 × 20 = 0.603 Volt
Average emf over a complete cycle is zero.
Maximum current in the coil
Imax =
εmax
R
 =
0.603
10
 = 0.0603 A
Average power lost =
1
2
εmax × Imax = 0.018 W
Source of the power is work done in rotating the coil.
© examsnet.com
Go to Question:
Prev Question
Next Question