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4.Moving Charges and Magnetism
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A proton moving with a velocity, $2.5 \times {10^7}\,m/s$, enters a magnetic field of intensity $2.5\,T$ making an angle ${30^o}$ with the magnetic field. The force on the proton is
A
$3 \times {10^{ - 12}}\,N$
B
$5 \times {10^{ - 12}}\,N$
C
$6 \times {10^{ - 12}}\,N$
D
$9 \times {10^{ - 12}}\,N$
Solution
$F = qvB\sin \theta = 1.6 \times {10^{ – 19}} \times 2.5 \times 2.5 \times {10^7}\sin {30^o}$
$F = 1.6 \times {10^{ – 19}} \times 6.25 \times {10^7} \times \frac{1}{2} = 5 \times {10^{ – 12}}\,N$
Standard 12
Physics
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