Question

A wire AB is carrying steady current 'I₁' and is kept on the table. Another wire CD carrying current 'I₂' is held parallel and directly above AB at a distance 'r'. When wire CD is left free and it remains suspended at its position, its mass per unit length is (g =acceleration due to gravity) ____________.

Options

• `(mu_0  "I"_2)/(2 pi"rg")`

• `(mu_0  "I"_1"I"_2)/(4 pi"rg")`

• `(mu_0  "I"_1)/(2 pi"rg")`

• `(mu_0  "I"_1"I"_2)/(2 pi"rg")`

Answer 1

A wire AB is carrying steady current 'I₁' and is kept on the table. Another wire CD carrying current 'I₂' is held parallel and directly above AB at a distance 'r'. When wire CD is left free and it remains suspended at its position, its mass per unit length is `(mu_0  "I"_1"I"_2)/(2 pi"rg")`.

Explanation:

Force per unit length between the current carrying wires is given by

`"F" = mu_0/(4pi) xx (2"I"_1 "I"_2)/"r" = "mg"`    (m = mass per unit length, r = distance between the wires)

`therefore "m" = (mu_0  "I"_1 "I"_2)/(2pi"rg")`