An α - particle travels in a circular path of radius 0.45, m in a magnetic field B = 1.2,Wb/{m^2} w

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4.Moving Charges and Magnetism

easy

An $\alpha - $ particle travels in a circular path of radius $0.45\, m$ in a magnetic field $B = 1.2\,Wb/{m^2}$ with a speed of $2.6 \times {10^7}\,m/\sec $. The period of revolution of the $\alpha - $ particle is

A

$1.1 \times {10^{ - 5}}\,\sec $

B

$1.1 \times {10^{ - 6}}\sec $

C

$1.1 \times {10^{ - 7}}\,\sec $

D

$1.1 \times {10^{ - 8}}\,\sec $

Solution

(c) $T = \frac{{2\pi m}}{{qB}} = \frac{{2\pi r}}{v} = \frac{{2 \times 3.14 \times 0.45}}{{2.6 \times {{10}^7}}} = 1.08 \times {10^{ – 7}}\sec $

Standard 12

Physics

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