A metal rod of length L is placed normal to a magnetic field B and rotated in a circular path with frequency f. The potential difference between it ends will be-

Engineering

Physics

Motional EMF

Question

πL²Bf

BL / f

πL²B / f

fBL

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Potential difference e,

$e = \frac{1}{2} {BωL}^{2}$

ω = 2πf

$e = \frac{1}{2} B \times 2 {πfL}^{2}$

e = πL²Bf

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