A charged particle of charge ' q ' is accelerated by a potential difference ' V ' enters a region of uniform magnetic field ' B ' at right angles to the direction of fiel

Question

A charged particle of charge ' q ' is accelerated by a potential difference ' V ' enters a region of uniform magnetic field ' B ' at right angles to the direction of field. The charged particle completes semicircle of radius ' r ' inside magnetic field. The mass of the charged particle is

TestHub · Accepted Answer

Kinetic energy gained by the charged particle when accelerated through a potential difference $V$ is $K = q V$ .

The momentum of the particle is $p = 2 m K = 2 m q V$ .

When the particle enters a uniform magnetic field $B$ perpendicularly, it moves in a circular path of radius $r$ given by $r = \frac{p}{qB}$ .

Substituting the value of $p$ , we get $r = \frac{2 m q V}{qB}$ .

Squaring both sides, $r^{2} = \frac{2 m q V}{q ^{2} B ^{2}} = \frac{2 mV}{q B ^{2}}$ .

Rearranging for mass $m$ , we get $m = \frac{r ^{2} q B ^{2}}{2 V}$ .

Answer: $\frac{r ^{2} q B ^{2}}{2 V}$

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