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Physics-

General

Easy

Question

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

$½ B ω l^{2}$
$3 / 4 B ω l^{2}$
$B ω l^{2}$
$2 B ω l^{2}$

The correct answer is: $½ blank B omega l to the power of 2 end exponent$

If in time $t$ , the rod turns by an angle $theta$ , the area generated by the rotation of rod will be $equals fraction numerator 1 over denominator 2 end fraction l cross times l theta equals fraction numerator 1 over denominator 2 end fraction l to the power of 2 end exponent theta$

So the flux linked with the area generated by the rotation of rod
$ϕ equals B open parentheses fraction numerator 1 over denominator 2 end fraction l to the power of 2 end exponent theta close parentheses cos invisible function application 0 equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent theta equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega t$
And so $e equals fraction numerator d ϕ over denominator d t end fraction equals fraction numerator d over denominator d t end fraction open parentheses fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega t close parentheses equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega$

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$A blank p a r t i c l e blank o f blank m a s s blank m blank i s blank m o v i n g blank i n blank a blank h o r i z o n t a l blank c i r c l e blank o f blank r a d i u s blank r comma blank u n d e r blank a blank c e n t r i p e t a l blank f o r c e$
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