$(A)\;\frac{K 2 \lambda_1 \lambda_2}{R} \\ (B)\;K \frac{2 \lambda_1^2}{R} \\ (C)\; Zero \\ (D)\;none $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Electric field due to wire (2)$= \large\frac{1}{4 \pi \in _0} \frac{2 \lambda_2}{R}$

Charge per unit length on wire $(1)=\lambda _2$

Force per unit length on wire $(1) =\lambda_1 \times E$

$\qquad= \lambda \times \large\frac{1}{4 \pi \in _0}\frac{2 \lambda_2}{R}$

$\qquad= \large\frac{K(2 \lambda _1 \lambda_2)}{R}$

Hence A is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...