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

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$\frac{μ_{0} i q v}{2 π a}$

$\frac{μ_{0} i q v}{4 π a}$

$\frac{μ_{0} i q v l}{2 π a}$

$\frac{μ_{0} i q v l}{4 π a}$

answer is A.

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Detailed Solution

$B = \frac{μ_{0}}{2 π} \frac{i}{a}$

$\begin{array}{rcl} F & = & B q v \sin 90 ° \\ = & \frac{μ_{0}}{2 π} \frac{i}{a} (q v) \end{array}$

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