A wire AB is carrying steady current I 1 and is kept on the table. Another wire CD carrying current I 2 is held directly above as shown in figure. When the wire CD is left free and it remains susp

Question

A wire AB is carrying steady current I 1 ​ and is kept on the table. Another wire CD carrying current I 2 ​ is held directly above as shown in figure. When the wire CD is left free and it remains suspended at its position, its mass per unit length is (g=acceleration due to gravity, μ 0 ​ = permeability of free space)

TestHub · Accepted Answer

The magnetic force per unit length between two parallel wires carrying currents $I_{1}$ and $I_{2}$ separated by a distance $r$ is given by:

$F_{m} = \frac{μ _{0} I _{1} I _{2}}{2 π r}$

Since the currents in wires AB and CD are in opposite directions, the magnetic force between them is repulsive. This upward force on wire CD balances its downward gravitational force.

Let $λ$ be the mass per unit length of wire CD. The weight per unit length is:

$F_{g} = λ g$

For wire CD to remain suspended in equilibrium, the magnetic force per unit length must equal the weight per unit length:

$F_{m} = F_{g}$

$\frac{μ _{0} I _{1} I _{2}}{2 π r} = λ g$

$λ = \frac{μ _{0} I _{1} I _{2}}{2 π r g}$

Answer: $\frac{μ _{0} I _{1} I _{2}}{2 π r g}$

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