First slide

Motional EMF

Question

A conducting rod of length 2l is rotating with constant angular speed $ω$ about its perpendicular bisector. A uniform magnetic field $\vec{B}$ exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is

Moderate

A

$B ω l^{2}$
B

$\frac{1}{2} {Bωl}^{2}$
C

$\frac{1}{8} {Bωl}^{2}$
D

Zero

Solution

Potential difference between O and A is
$V_{0} - V_{A} = \frac{1}{2} B l^{2} ω$
Potential difference between O and B is
$V_{o} - V_{B} = \frac{1}{2} B l^{2} ω$
$so V_{A} - V_{B} = 0$

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