Question

A metal rod $\text{PQ}$ is kept near an infinitely long, straight conductor, as shown. It carries a current $I$ and has a uniform area of cross - section $A$. If $J$ is the current density in the rod and $n$ is a number of electrons per unit volume, then the force on rod $\text{PQ}$ is

• A. $\frac{2{\mu }_{0}eJ{I}_1}{\pi nea}$
• B. $\frac{{\mu }_{0}eJ{I}_1}{\pi nea}$
• C. $\frac{3{\mu }_{0}eJ{I}_1}{2\pi nea}$
• D. $\frac{{\mu }_{0}eJ{I}_1}{2\pi nea}$

Answer 1

The correct option is D $\frac{{\mu }_{0}eJ{I}_1}{2\pi nea}$

Magnetic field at the rod $\text{PQ}$ due to infinite wire will be

$B=\frac{{\mu }_0{I}_1}{2\pi a}$


Now the force acting on an electron present in rod $\text{PQ}$, due to magnetic field of $I_1$ is

$F_m=evB(\because v\perp B)$

here $v=v_d$ (drift velocity of electron)

$\therefore F_m=Be{v}_d=\frac{{\mu }_{0}e{v}_d{I}_1}{2\pi a}......\left(1\right)$

Now, as we know

$I=nAe{v}_d$

$\therefore v_d=\frac{I}{nAe}=\frac{J}{ne}(\text{as}J=\frac{I}{A})$

So, equation $\left(1\right)$ becomes,

$F_m=\frac{{\mu }_{0}e{I}_1}{2\pi a}\frac{J}{ne}$

$\therefore F_m=\frac{{\mu }_{0}eJ{I}_1}{2\pi nea}$

Therefore, option $\left(D\right)$ is the correct answer.