Question

Obtain an expression for motional e.m.f. in a rotating bar with constant angular velocity ‘ω’ inuniform magnetic field perpendicular to plane of rotation.

Answer 1

Suppose a rod of length l is rotates anticlockwise, along an axis through one end and perpendicular to its length, in a plane perpendicular to a uniform magnetic field of induction `vecB`, as shown in figure; `vecB` points into the page. Let the rod’s constant angular speed be ω.

Consider an infinitesimal length element dr, distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is:

`(dA)/(dt)` = frequency of rotation × dA = f dA

where f = `ω/(2π)` is the frequency of rotation.

∴ `(dA)/(dt) = ω/(2πr  dr)`

= ω r dr

Therefore, the magnitude of the induced emf in the element is:

`|de| = (dΦ_m)/(dt)`

= `B (dA)/(dt)`

= B ω r dr

Since the emfs in all elements of the rod will be in sequence, the total emf induced across the ends of the revolving rod is:

`|e| = int de`

= `int_0^l B ω r dr`

= `Bω int_0^l r dr`

= `Bω l^2/2`