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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The magnetic force on the loop due to the field B is $BI l$ .

The magnetic force on the loop due to the field B is 2 $BI l$

The magnetic force on the loop due to the field B is $\frac{BI l R}{l + R}$

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.

$∴ F = {BI}_{1} l + {BI}_{2} l = BIl$

Where I₁ = current in part ADC

I_{2 =} current in part ABC

$I_{1} + I_{2} = 1$

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

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