First slide

Force on a charged particle moving in a magnetic field

Question

An electron moving with kinetic energy 6.6 x 10^-14 J enters a magnetic field 4 X10^-3 T at right angle to it, The radius of circular path will be nearest to

Moderate

A

25 cm
B

50 cm
C

75 cm
D

1m

Solution

$evB = {mv}^{2} / R or ebR = mv$
Squaring on both sides, we get
$\begin{array}{l} e^{2} B^{2} R^{2} = m^{2} v^{2} = 2 m [\frac{1}{2} {mv}^{2}] \\ = 2 m \times K.E. \\ ∴ R = \frac{(2 m \times K . E .)^{1 / 2}}{eB} \\ = \frac{{(2 \times 9 \cdot 1 \times 10^{- 31} \times 6 \cdot 6 \times 10^{- 14})}^{1 / 2}}{1 .6 \times 10^{- 19} \times 4 \times 10^{- 3}} \end{array}$
= 54 cm, which is nearest to 50 cm.

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