Question

A circular wire loop of radius ' $r$ ' carries a total charge '$Q$' distributed uniformly over its length. A small length dl of the wire is cut off. The electric field at the centre due to the remaining wire :-

• A. $\frac{\text{ Qd }\ell }{8{\pi }^2{\in }_0{r}^3}$
• B. $\frac{\text{ Qd }\ell }{2{\pi }^2{\in }_0{r}^3}$
• C. $\frac{\text{ Qd }\ell }{8\pi {\in }_0{r}^3}$
• D. $\frac{Qd\ell }{4{\pi }^2{\in }_0{r}^3}$

Answer 1

Answer: (A)

We know that electric field at the centre of the charged circular wire is zero. Here we may conclude that electric field due to the element of length $d\ell$ and the rest of the wire are equal and opposite. So we have merely to determine the electric field due to the charged element of length $d\ell$
$E=\frac{1}{4\pi {\in }_0}\frac{\lambda d\ell }{r^2}$
(where $\lambda =\frac{Q}{2\pi r})$
$=\frac{1}{4\pi {\in }_0}\frac{Q}{2\pi r}\frac{d\ell }{r^2}$
$=\frac{Qd\ell }{8{\pi }^2{\in }_0{r}^3}$