Question

A long horizontal wire "A" carries a current of 50 Amp. It is rigidly fixed. Another small wire "B" is placed just above and parallel to "A". The weight of wire-B per unit length is 75 × 10^–3 Newton/meter and carries a current of 25 Amp. Find the position of wire B from A so that wire B remains suspended due to magnetic repulsion. Also indicate the direction of current in B w.r.t. to A.

(a) (1/2) × 10^–2 m; in same direction

(b) (1/3) × 10^–2 m; in mutually opposite direction

(c) (1/4) × 10^–2 m; in same direction

(d) (1/5) × 10^–2 m; in mutually opposite direction

Answer 1

The correct option (b) (1/3) × 10^–2 m; in mutually opposite direction

Explanation:

Here weight of wire B is just supported by magnetic repulsion. Hence direction of current I₂ should be opposite to I₁.

Now    F_magnetic = F_{gravitational}

∴ (μ₀/2π)[(I₁I₂)/d] = w

∴  [(4π × 10^–7)/(2π)] × [(50 × 25)/d] = 75 × 10^–3

∴  [(2500 × 10^–7)/d] = 75 × 10^–3

∴  (1/d) = 300

∴  d = [1/(300)] = (1/3) × 10^–2 m

and direction of current in B is in opposite direction of current in A.