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Question

Physics Question on Electromagnetic induction

A metallic rod of length ll is tied to a string of length 2l2l and made to rotate with angular speed (ω\omega on a horizontal table with one end of the string fixed. If there is a vertical magnetic field BB in the region, the emf induced across the ends of the rod is

A

2Bωl32\frac{ 2 \, B \, \omega l^3 }{ 2 }

B

3Bωl32\frac{ 3 B \omega l^3 }{ 2 }

C

4Bωl22\frac{ 4 B \omega l^2 }{ 2 }

D

5Bωl22\frac{ 5 B \omega l^2 }{ 2}

Answer

5Bωl22\frac{ 5 B \omega l^2 }{ 2}

Explanation

Solution

e=2l3l(ωx)Bdx=Bω[(3l)2(2l)2]2e = \int \limits_{ 2 \, l }^{ 3 \, l } \, ( \omega \, x ) \, Bdx = B \omega \frac{ [ ( 3 \, l )^2 - ( 2 \, l)^2 ] }{ 2 }
= 5Bl2ω2\frac{ 5 B \, l^2 \, \omega }{ 2 }