A particle of mass m and charge q moves with a constant velocity v along positive x direction. It en

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4.Moving Charges and Magnetism

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A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

A

$qbB/m$

B

$q\left( {b - a} \right)\,B/\,m$

C

$qaB/\,m$

D

$q\left( {b + a} \right)\,B/\,m$

Solution

The radius $r$ of the circular path is given by (see Fig.)

$\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{qvB}$

or $v=\frac{q B}{m}(r)$

$\therefore \mathrm{v}_{\min }=\frac{\mathrm{qB}}{\mathrm{m}}\left(\mathrm{r}_{\min }\right)=\frac{\mathrm{qB}}{\mathrm{m}}(\mathrm{b}-\mathrm{a})$

which is choice $( 2)$

Standard 12

Physics

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