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A current carrying closed loop in the form of a trapezoid ABCDA is placed in a uniform magnetic field whose direction is perpendicular to AD and the field lies in the plane of the loop. If the magnitude of magnetic force experienced by AD is 10N, what is the magnetic force experienced by AB?

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4 N

3 N

4.5 N

2.5 N

answer is D.

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Detailed Solution

If $I$ be the current in the loop, Then

$I x 2 x B = 10 \Rightarrow IB = 5 A - T .$

Angle between AB and AD is $cosθ = \frac{0.5}{1} \Rightarrow θ = 60^{o}$

$∴$ Force experienced by AB $= I x 1 x B x \sin (90^{o} - 60^{o}) N$

$= 5 \cos 60^{o} N = 2.5 N$

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