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Q.

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

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a

1 $μs$

b

2 $μs$

c

0.5 $μs$

d

1.5 $μs$

answer is A.

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Detailed 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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