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Figure shows a conducting loop ABCDA placed in a uniform magnetic field (strength B) perpendicular to its plane. The part ABC is the (three-fourth) portion of the square of side length l. The part ADC is a circular arc of radius R. The points A and C are connected to a, battery which supplies a current i to the circuit.The magnetic force on the loop due to the field B is

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zero

BII

2BIR

BIlR/1+R

answer is B.

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Detailed Solution

$F = B i_{1} l + B i_{2} l$

$\Rightarrow F = B l (i_{1} + i_{2}) = B i l$

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