Q. A proton moving with a velocity $2.5 \times 10^{7}\ m/s$, enters a magnetic field of intensity $2.5T$ making an angle $30{^\circ}$ 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) $9 \times 10^{- 12}N$
Explanation
(b) $F = qvB\sin\theta = 1.6 \times 10^{- 19} \times 2.5 \times 2.5 \times 10^{7}\sin{30{^\circ}}$
$F = 1.6 \times 10^{- 19} \times 6.25 \times 10^{7} \times \frac{1}{2} = 5 \times 10^{- 12}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.