A square loop PQRS carrying a current of $6 A$ is placed near a long wire carrying current $10 A$ shown in the below figure. The magnetic force on the square loop is [[1]] $\times 1 0^{- 4} N$ .

Question

A square loop PQRS carrying a current of 6 A is placed near a long wire carrying current 10 A shown in the below figure. The magnetic force on the square loop is [[1]]×10-4 N.

Accepted Answer

A square loop PQRS carrying a current of

6 A

is placed near a long wire carrying current

10 A

shown in the below figure. The magnetic force on the square loop is

- 1.6 \times 1 0^{- 4} N

.

Magnetic field on SP is,

B_{1} = \frac{μ_{0} \times 10}{2 π \times 1}

= \frac{5 μ_{0}}{π}

The direction of current in the wire is opposite to the direction of current on the side SP of the square loop. Therefore the magnitude of current is given by,

F_{1} = B_{1} \times 6 \times 2

= 12 \times \frac{5 μ_{0}}{π}

= \frac{60 μ_{0}}{π}

Magnetic field on QR is,

B_{2} = \frac{μ_{0} \times 10}{2 π \times 3}

= \frac{5 μ_{0}}{3 π}

The direction of current in the wire is opposite to the direction of current on the side QR of the square loop. Therefore the magnitude of current is given by,

F_{2} = B_{2} \times 6 \times 2

= 12 \times \frac{5 μ_{0}}{3 π}

= \frac{20 μ_{0}}{π}

Now, RS and PQ are perpendicular to the straight line conductor, carrying current in opposite directions to each other so the net force on both of them will cancel each other.
Therefore net force on the square loop is

F = \frac{20 μ_{0}}{π} - \frac{60 μ_{0}}{π}

= - 40 \times 4 \times 1 0^{- 7}

= - 1.6 \times 1 0^{- 4} N

.
Here the negative sign represents that there is repulsion between the loop and the straight conductor.

A square loop PQRS carrying a current of $6 A$ is placed near a long wire carrying current $10 A$ shown in the below figure. The magnetic force on the square loop is [[1]] $\times 1 0^{- 4} N$ .

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