A 2.0 m long metallic rod is rotated with an angular frequency of 100 rad/s about an axis normal to the rod and passing through its one end. A uniform magnetic field of 2.0 T exits parallel to the axis. The emf induced between the two ends of the rod is:

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This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 2 Official Paper

Answer 1

Concept:

When a metallic rod moves in magnetic field then emf is induced in the across end of the rod
Induced emf, e = Blv
where, B = magnetic field, l = length of the rod, v is the velocity of the rod.

Calculation:

Length of the rod, l = 2 m, Angular frequency, ω = 100 rad/s, Magnetic field strength, B = 2 T

One end of the rod has zero linear velocity,

while the other end has a linear velocity of lω.

Average linear velocity of the rod, \(v = \frac {l\omega - 0 }{2}= \frac {2\times 100}{2}\) = 100 m/s

Induced emf, e = Blv = 2 × 2 × 100 = 400 V