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Question

Physics Question on Electric Current

A rigid square loop of side 'a' and carrying current I2_2 is lying on a horizontal surface near a long current I1_1 carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be :

A

Attractive and equal to μ0I1I23π\frac{\mu_0 I_1 I_2}{3 \pi}

B

Repulsive and equal to μ0I1I24π\frac{\mu_0 I_1 I_2}{4 \pi}

C

Repulsive and equal to μ0I1I22π\frac{\mu_0 I_1 I_2}{2 \pi}

D

Zero

Answer

Repulsive and equal to μ0I1I24π\frac{\mu_0 I_1 I_2}{4 \pi}

Explanation

Solution

F3_3 & F4_4 cancel each other Force on PQ will be F1=I2B1_1 = I_2 B_1 a =I2μ0I12πaa \, \, \, \, \, \, \, \, \, \, \, =I_2 \frac{\mu_0 I_1}{2 \pi a}a =μ0I1I22π \, \, \, \, \, \, \, \, \, \, \, = \frac{\mu_0 I_1 I_2}{2\pi} Force on RS will be F2_2 = I2B2_2 B_2 a =I2μ0I12π2aa \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = I_2 \frac{\mu_0 I_1}{2\pi 2a}a =μ0I1I24π\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{\mu_0 I_1 I_2}{4 \pi} Net force = F1F2=μ0I1I24πF_1 - F_2 \, = \, \frac{\mu_0 I_1 I_2}{4 \pi} repulsion Option (2)