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4.Moving Charges and Magnetism
easy
An electron (mass = $9.0 × $${10^{ - 31}}$ $kg$ and charge =$1.6 \times {10^{ - 19}}$ $coulomb$) is moving in a circular orbit in a magnetic field of $1.0 \times {10^{ - 4}}\,weber/{m^2}.$ Its period of revolution is
A
$3.5 \times {10^{ - 7}}$ $sec$
B
$7.0 \times {10^{ - 7}}$ $sec$
C
$1.05 \times {10^{ - 6}}$ $sec$
D
$2.1 \times {10^{ - 6}}$ $sec$
Solution
(a) $T = \frac{{2\pi m}}{{qB}} = \frac{{2 \times 3.14 \times 9 \times {{10}^{ – 31}}}}{{1.6 \times {{10}^{ – 19}} \times 1 \times {{10}^{ – 4}}}} = 3.5 \times {10^{ – 7}}$ $sec$
Standard 12
Physics
