First slide

Motional EMF

Question

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity $ω .$ The induced e.m.f. between the two ends is

Moderate

A

$\frac{1}{2} {Bωl}^{2}$
B

$\frac{3}{4} {Bωl}^{2}$
C

${Bωl}^{2}$
D

$2 {Bωl}^{2}$

Solution

If in time the rod turns by an angle $θ,$ the area generated by the rotation of rod will be
$= \frac{1}{2} l \times lθ = \frac{1}{2} l^{2} θ$
So the flux linked with the area generated by the rotation of rod
$ϕ = B (\frac{1}{2} l^{2} θ) \cos 0 ° = \frac{1}{2} {Bl}^{2} θ = \frac{1}{2} {Bl}^{2} ωt$
$So, e = \frac{dϕ}{dt} = \frac{d}{dt} (\frac{1}{2} {Bl}^{2} ωt) = \frac{1}{2} {Bl}^{2} ω$

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