Question

In the figure shown, the wires $\text{AB}$ and $\text{PQ}$ carry steady currents, $I_1$ and $I_2$ respectively. $\text{PQ}$ is of uniformly distributed mass $m$ and length $l$. $\text{AB}$ and $\text{PQ}$ are both horizontal and kept in the same vertical plane. The $\text{PQ}$ is in equilibrium at height $h$. Find $h$ in terms of $I_1,I_2,l,m$ and other standard constants.

• A. $\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi mg}$
• B. $\frac{{\mu }_0{I}_1{I}_{2}l}{\pi mg}$
• C. $\frac{{\mu }_0{I}_1{I}_{2}l}{4\pi mg}$
• D. $\frac{2{\mu }_0{I}_1{I}_{2}l}{\pi mg}$

Answer 1

The correct option is A $\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi mg}$

As the currents are in opposite direction, so the magnetic repulsive force experienced by the wire $\text{PQ}$ is,

$F_B=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi h}\left[\text{In the upward direction}\right]$


Since the wire $PQ$ is in equilibrium,

$\therefore F_B=mg$

$\Rightarrow \frac{{\mu }_0{I}_1{I}_{2}l}{2\pi h}=mg$

$\Rightarrow h=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi mg}$

Hence, option (a) is the correct answer.