Question

A horizontal wire AB of length ''l'' and mass "m" carries a steady current I₁ , free to move in vertical plane is in equilibrium at a height of "h" over another parallel long wire CD carrying a steady current I₂ , which is fixed in a horizontal plane. Derive the expression for the force acting per unit length on the wire AB and write the horizontal condition for which wire AB is in equilibrium.

Answer 1

Given :

Length of the horizontal wire AB = $l$

mass of wire AB = m

The steady current carried by wire AB = $I_1$

The steady current carried by wire CD =$I_2$

Wire AB is free to move in vertical plane and CD is fixed in a horizontal plane .

The height from AB above which CD is placed = h

To Derive :

An expression for the force acting per unit length on the wire AB .

To Find :

The  condition for which wire AB is in horizontal equilibrium  = ?

Solution:

Consider a  small element dl of the wire CD, the magnetic field at 'dl' due to the wire AB is :

$B= \frac{\mu_0I_1}{2 \pi h}$       - (1)

The field due to the portions of the wire CD , above and below dl is zero .

Thus equation (1) gives out field at dl.

The direction of this field is perpendicular to the plane of the diagram and going into it.

The magnetic force at element dl is :

$d\vec F= I_2 \vec {dl}\times \vec B$    towards  AB

Now , force per unit length of CD on AB is :

$\frac{dF}{dl}=\frac{\mu_0 I_1 I_2}{2 \pi h}$

The condition for AB to be in equilibrium :

$\frac{\mu_0 I_1I_2}{2 \pi h}= \frac{mg}{l}$

Answer 2

Explanation:

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