Question

Figure 6 below shows a long straight conductor X carrying a current I₁. A point P is at a perpendicular distance ‘r₁’ from it.

1. How much force conductor X exerts on a short conductor Y of length ‘l’ which is carrying a current ‘I’ and is kept at point P parallel to the conductor X?
2. Another long wire Z carrying a current I₂ is now kept parallel to X and Y at a distance ‘r₂’ from Y such that the conductor Y remains at rest. Obtain the relation between the currents I₁ and I₂.

Answer 1

(i) Force exerted by conductor X on conductor Y

The magnetic field B₁ produced by a long straight conductor X carrying current I₁ at a perpendicular distance r₁ is given by:

`B_1 = (mu_0 I_1)/(2pir_1)`

The force F_x acting on conductor Y(of length l and current I) placed in this magnetic field is:

F_X = B₁Il sin(θ)

Since the conductors are parallel, the magnetic field is perpendicular to conductor Y(θ = 90°, so sin 90° = 1). Substituting the value of B₁.

`F_X = (mu_0I_1Il)/(2pir_1)`

Since both currents I₁ and I are in the same direction (upwards), the force is attractive, meaning conductor Y is pulled towards conductor X.

(ii) Relation between currents I₁ and I₂ for equilibrium

For conductor Y to remain at rest, the net force on it must be zero. This means the force exerted by conductor Z (F_Z) must be equal in magnitude and opposite in direction to F_X.

Force from Z: Similar to F_X, the force exerted by conductor Z on Y at distance r₂ is:

`F_Z = (mu_0I_2Il)/(2pir_2)`

Since I₁ and I₂ are in the same direction, this force is also attractive, pulling Y towards Z (to the right).

Equilibrium Condition: F_X = F_Z

`(mu_0I_1Il)/(2pir_1) = (mu_0I_2Il)/(2pir_2)`

`(I_1)/(r_1) = (I_2)/(r_2)`

I₁r₂ = I₂r₁

or `(I_1)/(I_2) = (r_1)/(r_2)`