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Standard XII

Physics

Law of Radioactivity

The loop is s...

Question

The loop is shown moves with a velocity v in a uniform magnetic field of magnitude B, directed into the paper. The potential difference between P and Q is e. Then:

e = \frac{1}{2} B l v

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e = B l v

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P is negative with respect to Q

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Q is negative with respect to P

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Solution

The correct option is A $e = \frac{1}{2} B l v$
Given: potential difference =e, velocity=v, B=into the the plane of plane of paper.
Solution: As we know the formula of the formula of motional emf
$ε = B v l$
But here $l = \frac{l}{2}$
So, $ε = \frac{B v l}{2}$
Hence the correct answer is A

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The loop is shown moves with a velocity v in a uniform magnetic field of magnitude B, directed into the paper. The potential difference between P and Q is e. Then:

The correct option is A e=12BlvGiven: potential difference =e, velocity=v, B=into the the plane of plane of paper.Solution: As we know the formula of the formula of motional emf ε=BvlBut here l=l2So,ε=Bvl2Hence the correct answer is A

The correct option is A $e = \frac{1}{2} B l v$
Given: potential difference =e, velocity=v, B=into the the plane of plane of paper.
Solution: As we know the formula of the formula of motional emf
$ε = B v l$
But here $l = \frac{l}{2}$
So, $ε = \frac{B v l}{2}$
Hence the correct answer is A