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Standard XII

Physics

Cyclotron

A particle of...

Question

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

\frac{q b B}{m}

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\frac{q (b - a) B}{m}

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\frac{q a B}{m}

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\frac{q (b + a) B}{2 m}

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Solution

The correct option is B $\frac{q (b - a) B}{m}$
Radius of curvature has to be greater than the width of magnetic field for the particle to enter the region $x > B$ .

$∴ \frac{m v}{q B} > (b - a)$

$\Rightarrow v > \frac{q B (b - a)}{m}$

$∴ v_{m i n} = \frac{q B (b - a)}{m}$

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The correct option is B q(b−a)BmRadius of curvature has to be greater than the width of magnetic field for the particle to enter the region x>B.∴mvqB>(b−a)⇒v>qB(b−a)m∴vmin=qB(b−a)m

The correct option is B $\frac{q (b - a) B}{m}$
Radius of curvature has to be greater than the width of magnetic field for the particle to enter the region $x > B$ .

$∴ \frac{m v}{q B} > (b - a)$

$\Rightarrow v > \frac{q B (b - a)}{m}$

$∴ v_{m i n} = \frac{q B (b - a)}{m}$