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4.Moving Charges and Magnetism
medium
A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should be
A
$6.5 \times {10^{15}}\,m/{\sec ^2}$
B
$6.5 \times {10^{13}}\,m/{\sec ^2}$
C
$6.5 \times {10^{11}}\,m/{\sec ^2}$
D
$6.5 \times {10^9}\,m/{\sec ^2}$
Solution
(a) $F = ma = qvB$ $==>$ $a = \frac{{qvB}}{m} = \frac{{1.6 \times {{10}^{ – 19}} \times 2 \times 3.4 \times {{10}^7}}}{{1.67 \times {{10}^{ – 27}}}}$
$=$ $6.5 \times {10^{15}}\,m/{\sec ^2}$
Standard 12
Physics
