Question

A rectangular loop of sides 25 cm and 10 cm carrying current of 15 A is placed with its longer side parallel to a long straight conductor 2.0 cm apart carrying a current of 25 A. What is the new force on the loop?

Solution

When the loop is placed with the longest side parallel to the long straight wire, then the shortest sides of the loop at the same distance from the wire and force on them are equal but in opposite direction therefor net force on two short side are zero.Now longest side which near the wire, force on it is given by $$F_1=B_1I_{loop}L$$whereB is the magnetic field due to wire L is the length of the side of the loop$$B=\dfrac{\mu_0I}{2\pi r}$$where r is the distance of the sides of the loopLet direction of the current in wire is in upward direction.$$F_1=\dfrac{\mu_0I_{wire}I_{loop}}{2\pi r}$$$$F_1=\dfrac{4\pi\times10^{-7}\times25\times15\times25}{2\pi \times2}$$$$F_1=9.37\times10^{-4}\ N$$ away from the wire.Force on the longest side which is away from the wire$$F_2=\dfrac{\mu_0I_{wire}I_{loop}}{2\pi (r+length\ of\ the\ shortest side}$$$$F_2=\dfrac{4\pi\times10^{-7}\times25\times15\times25}{2\pi\times12}$$$$F_2=1.56\times10^{-4}\ N$$ towards from the wire.Net force is given by$$F=F_1-F_2=7.81\times10^{-4}$$ away from the wirePhysicsNCERTStandard XII

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