First slide

Force on a charged particle moving in a magnetic field

Question

A magnetic field of induction 8= 35.34x10^-6T is applied on an electron in a direction perpendicular to its motion. Find the time required for the electron to complete one revolution. Assume its mass and charge

Moderate

A

1 $μs$
B

2 $μs$
C

0.5 $μs$
D

1.5 $μs$

Solution

Force due to magnetic field = B e v
Normal force on electron $= \frac{{mV}^{2}}{r}$
$∴ Bev = \frac{{mv}^{2}}{r}$
or $v = \frac{Ber}{m}$
The time for one revolution is
$\begin{array}{l} T = \frac{2 πr}{V} = \frac{2 πrm}{Ber} = \frac{2 πm}{Be} \\ = \frac{2 \times 3 \cdot 14 \times (9 \cdot 1 \times 10^{- 31})}{(35 \cdot 34 \times 10^{- 19}) (1 \cdot 6 \times 10^{- 19})} \\ = 1 \times 10^{- 6} \sec = 1 μs \end{array}$

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