Q.
A semicircular wire of radius $R$ is rotated with constant angular velocity about an axis passing through one end and perpendicular to the plane of the wire. There is a uniform magnetic field of strength $B$. The induced emf between the ends is
Electromagnetic Induction
Solution:
We connect a conducting wire from $A$ to $C$ and complete the semicircular loop.
The emf in the semicircular loop is zero because its magnetic flux does not change.
$\therefore$ emf of section $A P C+$ emf of section $C Q A=0$
$\therefore $ emf of section $A P C=$ emf of section $A Q C=2 B R^{2} \omega$
