Gujarati
4.Moving Charges and Magnetism
easy

A electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at right angle to the uniform magnetic field $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete a circular orbit is......$µs$

A

$2$

B

$4$

C

$3$

D

$1$

Solution

(d) $T = \frac{{2\pi m}}{{qB}} = \frac{{2 \times 3.14 \times 9.1 \times {{10}^{ – 31}}}}{{1.6 \times {{10}^{ – 19}} \times 3.534 \times {{10}^{ – 5}}}}$
$ = 1 \times {10^{ – 6}}\,sec = 1\,\mu sec.$

Standard 12
Physics

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