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Q.

The figure shows a conducting loop ABCDA placed in a uniform magnetic field perpendicular to its plane. The part ABC is the (3/4)th 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 supply a current I to the circuit. The magnetic force on the loop due to the field B is

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a

The magnetic force on the loop due to the field B is 2BIl

b

The magnetic force on the loop due to the field B is BIl.

c

The magnetic force on the loop due to the field B is BIlRl+R

d

The magnetic field due to the loop is zero everywhere outside it.

answer is A.

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Detailed Solution

For finding the force, the circular part ADC can be replaced with a straight conductor AC.
Similarly, the part ABC can also be replaced with a straight conductor AC.

F=BI1l+BI2l=BIl

Where I1 = current in part ADC

I2 = current in part ABC

I1+I2=1

We cannot say that the field due to the loop is zero at all outside points.

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The figure shows a conducting loop ABCDA placed in a uniform magnetic field perpendicular to its plane. The part ABC is the (3/4)th 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 supply a current I to the circuit. The magnetic force on the loop due to the field B is