Question

A long wire $\text{AB}$ is placed on a table. Another wire $\text{PQ}$ of mass $1.0\text{g}$ and length $50\text{cm}$ is set to slide between two parallel rails as shown in the figure. A current of $50\text{A}$ is passed through the wires. At what distance above $\text{AB}$, will the wire $\text{PQ}$ be in equilibrium, if the entire setup is placed vertically, perpendicular to the plane of table, with $\text{AB}$ lying on the table.

• A. $25\text{mm}$
• B. $50\text{mm}$
• C. $75\text{mm}$
• D. $100\text{mm}$

Answer 1

The correct option is A $25\text{mm}$

Let the wire $\text{PQ}$ be at a distance $d$ from $\text{AB}$, when in equilibrium.

Weight of $\text{PQ}$ $=$ Force on $\text{PQ}$ due to magnetic field of $\text{AB}$

$\Rightarrow mg=F_m$

Now, $F_m=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}.l_{PQ}$

$F_m=\frac{4\pi \times {10}^{-7}\times 50\times 50}{2\pi \times d}\times 0.5=\frac{2.5\times {10}^{-4}}{d}$

$F_m=\frac{2500}{d}\text{N}$

$mg={10}^{-3}\times 10={10}^{-2}\text{N}$

$\therefore {10}^{-2}=\frac{2.5\times {10}^{-4}}{d}$

$\therefore d=2.5\times {10}^{-2}\text{m}=25\text{mm}$

Why this Question? These type of questions are typically called Rail-problems. These type of problems more often than not requires estabilishing the conditions of equilibrium. In this particular problem, $W=F_{mag}$