NEET | 2016Current Electricity and Electrical InstrumentEasy

Question

A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XYcarrying a current I, the net force on the loop will be:

Options

A.

$\frac{2 μ_{0} I i}{3 π}$

B.

$\frac{μ_{0} I i}{2 π}$

C.

$\frac{2 μ_{0} I i L}{3 π}$

D.

$\frac{μ_{0} I i L}{2 π}$

Solution

$F_{1} = \frac{μ_{0} I i L}{2 π} = \frac{μ_{0} I i}{π}$
$2 = \frac{μ_{0} I i L}{2 π \frac{3 L}{2}} = \frac{μ_{0} I i}{3 π}$
$∴$ F_net = F1 - F2
F_net = $\frac{2}{3} \frac{μ_{0} I i}{π}$

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