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Circular Kinematics

An electron h...

Question

An electron having momentum 2.4 X $10^{- 23}$ kg m $s^{- 1}$ enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of $30^{0}$ with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be :

A

2 mm

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B

1 mm

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C

\frac{\sqrt{3}}{2}

mm

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D

0.5 mm

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Solution

The correct option is C 0.5 mm
The radius of the helical path of the electron in the uniform magnetic field is
$r = \frac{m v_{⊥}}{e B} = \frac{m v s i n θ}{e B} = \frac{(2.4 \times 10^{- 23} k g m s^{- 1}) \times s i n 30^{o}}{(1.6 \times 10^{- 19} C) \times (0.15 T)}$
$= 5 \times 10^{- 4} m = 0.5 \times 10^{- 3} m = 0.5 m m$

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