JEE Main 2019 — Physics solution

JEE Main2019Physics

A rigid square loop of side ‘a’ and carrying current $l_{2}$ is lying on a horizontal surface near a long current $l_{1}$ carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be :

A
Repulsive and equal to $\frac{μ _{0} l _{1} l _{2}}{4 π}$
B
Repulsive and equal to $\frac{μ _{0} l _{1} l _{2}}{2 π}$
C
Zero
D
Attractive and equal to $\frac{μ _{0} l _{1} l _{2}}{3 π}$

Solution

$F = l_{2} a$ $(B_{1} - B_{2})$

$B_{1} = \frac{μ _{0} l _{1}}{2 π a}$

$B_{2} = \frac{μ _{0} l _{1}}{4 π a}$

$F = \frac{μ _{0} l _{1} l _{2}}{4 π}$

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