4.Moving Charges and Magnetism
medium

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.