Two long, straight, parallel wires of length 4.5 m carry parallel currents of 3.

Question

Two long, straight, parallel wires of length 4.5 m carry parallel currents of 3.1 A and 1.1 A. (a) If the wires are separated by a distance of 3.1 cm, what is the magnitude of the force between the two wires? N (b) Is this force attractive or repulsive? attractive repulsive (c) If the currents are in opposite directions (antiparallel), how do the answers to parts (a) and (b) change? What is the magnitude of the force between the two wires? N Is this force attractive or repulsive? attractive. repulsive.

Explanation / Answer

Magnetic force between two current carrying wires A and B is given by Fba = ( oIaIb/2d )L

Where Ia and Ib are the currents through them, and L is the length of those wires and d is the distance of separation and µ0 = 4 × 107 H·m1

L = 3.1 cm = 0.031 m

So, Fba = µ0 = [ 4 × 107 x 3.1 x 1.1 / (2 x 0.031) ] x 4.5

=> F = 9.9 x 10-5 N

Since it is given that the currents are parallel, that is they are in the same direction, this force is attractive.

If the currents are in opposite direction (anti-parallel), then the magnitude of this magnetic force between them will not change but the force now will be repulsive.

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