Question

A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0 × 10−4 kg m−1 is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

Answer 1

Given:

Magnitude of current, i1 = 10 A

Separation between two wires, d = 5 mm

Linear mass density of the second wire, λ = 1.0 × 10−4 kgm−1

Now,

Let i2 be the current in the second wire in opposite direction.

Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by

 Fml = μ0i1i22πd  (upwards)

Also,

Weight of the second wire, W = mg

Weight per unit length of the second wire, Wl = λg (downwards)

Now, according to the question,

Fml = Wl⇒μ0i1i22πd=λg

⇒2×10−7×50×i25×10−3 = 1×10−4×9.8⇒i2=9.8×10−720×10−7 = 0.49 A

Answer 2

Answer:

this is a your answer to the question answer