Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A conducting rod of mass $m$ and length $\ell$ is placed over a smooth table. A uniform magnetic field is acting perpendicular to the rod. Charge $q$ is suddenly passed through the rod and it acquires a velocity $v$ on the surface, then $q$ is equal to

$\begin {array} {1 1} (a)\;\large\frac{2mv}{B\ell} & \quad (b)\;\large\frac{mv}{B\ell} \\ (c)\;\large\frac{B\ell}{mv} & \quad (d)\;\large\frac{B\ell}{2mv} \end {array}$


Can you answer this question?

1 Answer

0 votes
Impulse = Change in linear momentum
$\int F\: dt = \int \: (i\ell B)\: dt = mv$
$\int \: i\: dt = q$
$mv = B\ell q$
Ans : (b)
answered Feb 22, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App