CBSE Class 12 · 2012 · Electromagnetic InductionShort Answer

A metallic rod of length

L

is rotated with angular frequency

ω

with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius

L

, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field

B

, parallel to the axis, is present everywhere. Deduce the expression for the emf induced between the centre and the metallic ring.

Answer: $ε = \frac{1}{2} B ω L^{2}$ .

Solution

Consider an element of the rod of length

d r

at distance

r

from the centre. It moves with speed

v = ω r

perpendicular to

B

, so the motional emf across it is

d ε = B v d r = B ω r d r

Adding the contributions from centre (

r = 0

) to rim (

r = L

ε = \int_{0}^{L} B ω r d r = B ω \frac{L ^{2}}{2}

ε = \frac{1}{2} B ω L^{2}

CBSE Class 12 Board · 2012 · Subjective