Question

A conducting rod of length L is rotated in a magnetic field B with an angular velocity ω, so that the rotational plane of rod is perpendicular to the magnetic field Determine the induced emf between the ends of the rod.

Answer 1

Induced emf in a Metal Rod Rotating in a Uniform Magnetic Field:

In figure, a uniform magnetic field is shown by cross (×) whose direction is normally inwards the surface. A conducting rod OA of length L rotates in the anticlockwise direction in magnetic field B with angular velocity ω. The plane of rotation of rod is perpendicular to the magnetic field. When an element dl of rod moves with velocity v normally with the magnetic field, the induced emf of the element.

If the distance between small element and centre is l, then
v = ωl
Thus, dε = Bωldl
Thus, to determine induced emf in the rod, we integrate the above equation from zero to L,

Using Fleming’s right hand rule, the direction of induced current in rod is from A to O. Thus, the end O of the rod is positive and A is negative.
If the frequency of rotation of rod is f, then to
ω = 2πf
Thus, ε = \(\frac{1}{2}\)B × 2πf × L²
= B × πL² × f
If the area covered by the rod in magnetic field is represented by A then
πL² = A
then ε = BAf …………………. (2)
Let the resistance of rod ‘R’ then