Question

rectangular coil of sides 0.25m & 0.1m carrying a current 12A is placed with its longer side parallel to a long straight conductor 0.02m apart carrying a current of 20A. Calculate the net force on the current loop​

Answer 1

Rectangular coil of sides 0.25 m and 0.1 m carrying a current 12 A is placed with its longer side parallel to a long straight conductor 0.02m apart carrying a current of 20A.

We have to calculate the net force on the current loop.

see the diagram attached in the answer, PQRS is the rectangular loop of sides 25cm or 0.25m and 10cm or 0.10m . and X is a straight long wire is placed 2cm or 0.02m away from the loop.

here it is obvious that F₃ and F₄ are equal and opposite because currents flow through them are equal and opposite and hence, net force due to these sides (SP and RQ) = 0.

now we have to find the force due to the remaining two sides on wire.

we know, force acting by two straight parallel wires is given by,

$F=\frac{\mu i_1i_2l}{2\pi r}$

where l is length of each parallel wire and r is the distance between them.

force due to PQ side, currents flow through PQ and X wires are in the opposite direction so the force must be repulsive.

and its magnitude is..

$F=\frac{4\pi\times10^{-7}\times12\times15\times0.25}{2\times3.14\times0.02}$

= 9.375 × 10¯⁴ N (repulsive)

similarly, force due to RS, currents flow through RS and X wires are in the same direction so the force acting on the wire is attractive.

and its magnitude is..

$F=\frac{4\pi\times10^{-7}\times12\times15\times0.25}{2\times3.14\times(0.02+0.1)}$

= 1.563 × 10¯⁴ N (attractive)

now net force acting on the wire = F₁ + F₂

= 9.375 × 10¯⁴ N (repulsive) + 1.563 × 10¯⁴ N (attractive)

= 7.812 × 10¯⁴ N (repulsive).

Therefore the force acting on the current loop is 7.812 × 10¯⁴ N