# A square loop of side 20 cm carrying current of 1A is kept near an infinite long straight wire carrying a current of 2A in the same plane as shown in the figure. Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor.

Consider current flowing through wire is I

_{1}and current flowing through loop is I

_{2}

_{.}Let c is the distance between loop and wire and a is side of square loop.

The current carrying long wire produces a magnetic field.

Using the right-hand rule 2 it can be determined that the magnetic field B is perpendicular to the plane of the loop and directed into the page.

Now force exerted by long wire on the two perpendicular side of loop are opposite in direction due to symmetry and hence cancels out.

Thus the magnetic force exerted on the entire loop due to long wire is because of the interaction of the sides parallel to the long wire.

Now these parallel side of loop acts like two wires carrying current one in same direction and one in opposite direction with current carrying long wire.

So by right hand rule we can predict that the direction of the force exerted on the side closer to the long wire is attractive (with side having current in same direction with long wire) and the force exerted on the farther side is repulsive (with side having current in opposite direction with long wire).

Hence the magnetic force exerted on the entire wire is toward the long wire and the magnitude of the magnetic force can be given as

$F={I}_{2}a\left(\frac{{\mu}_{o}{I}_{1}}{2\pi a}\right).\mathrm{sin}{90}^{o}-{I}_{2}a\left(\frac{{\mu}_{o}{I}_{1}}{2\pi (c+a)}\right).sin{90}^{o}\phantom{\rule{0ex}{0ex}}F=\frac{{\mu}_{o}{I}_{1}{I}_{2}a}{2\pi}\left(\frac{1}{a}-\frac{1}{c+a}\right)\phantom{\rule{0ex}{0ex}}F=\frac{\left(4\pi \times {10}^{-7}N/{A}^{2}\right).1A.2A.0.2m}{2\pi}\left(\frac{1}{0.2m}-\frac{1}{0.3m}\right)\phantom{\rule{0ex}{0ex}}F=1.33\times {10}^{-7}N$

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