Question

A square loop PQRS carrying a current of 6.0 A is placed near a
long wire carrying 10 A as shown in the figure. (a) Show that the
magnetic force acting on the part PQ is equal and opposite to
that on the part RS. (b) Find the magnetic force on the square
loop.
2 cm
R
10A
6A
1 cm​

Answer 1

Explanation:

First thing to note is the magentic field of the longer wire is directed into the plane of the paper, and since it is very long, its magnitude is inversely proportional to the distance (r) from the wire, given by a simple application of Ampere's Law (standard formula):

B=μ0I2πr.

Now since it is into the plane of the paper, and the Lorentz force is given by F=qv×B, the total force on the arms PQ and SR will cancel out, being equal in magnitude and opposite in direction (one can imagine a small element on PQ and an identical twin at the same distance on SR: the forces on them balance: this is true for all such small elements at different distances).

Now, one can similarly work out the forces on the other arms, which are although opposite in direction but not equal in magnitude. The force on a line element of a wire being dF=Idl×B, the net force is I2aμ0I2π(d−1−(a+d)−1).

Here I have used I2 as the current on the square, I for the current on the longer wire, d for the distance given, a for the side of the square, and SI units. Convert the units correctly and put the numbers to get the force in newton.

Good luck