Q. Electric charge q is distributed uniformly over a rod of length *l*. The rod is placed parallel to a long wire carrying a current *i*. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity *v* is

(a) $ \displaystyle \frac{\mu_0 i q v}{2 \pi a} $

(b) (a) $ \displaystyle \frac{\mu_0 i q v}{4 \pi a} $

(c) (a) $ \displaystyle \frac{\mu_0 i q v l}{2 \pi a} $

(d) (a) $ \displaystyle \frac{\mu_0 i q v l}{4 \pi a} $

Ans: (a)

Sol: Magnetic field at a distance ‘ a ‘ due to long wire is

$ \displaystyle B = \frac{\mu_0}{4\pi} \frac{2i}{a} $

Force acting on the rod is $\displaystyle F = q v B $ (Since velocity v & Mag. field B are perpendicular )

$ \displaystyle F = q v (\frac{\mu_0}{4\pi} \frac{2i}{a}) $

$\displaystyle F = \frac{\mu_0 q v i}{2 \pi a} $