# Copper disc of radius 0.1 m is rotated about its centre with 20 rev/s in a uniform magnetic field. EMF INDUCED across disc is pi/20 V, pi/10V, 20 pi mV, 10 pi mV

Sumit Majumdar IIT Delhi
7 years ago
Dear student,

The emf in this case would be given by:

Regards
Sumit
ankit singh
2 years ago
explanation : use formula, \xi_{in}=\frac{1}{2}B\omega R^2
where \xi_{in} denotes induced emf, B is magnetic field, \omega denotes angular frequency of disc and R denotes the radius of disc.
here, B = 0.1 T , R = 0.1 m and \omega=10rps=20π rad/s
hence, \xi_{in} = 1/2 × 0.1 × 20π × (0.1)²
= 1/2 × 20π × 0.01
= π/10 V