Q. A proton of energy $200\ MeV$ enters the magnetic field of $5\ T$. If direction of field is from south to north and motion is upward, the force acting on it will be
A) Zero
B) $1.6 \times 10^{- 10}N$
C) $3.2 \times 10^{- 8}N$
D) $1.6 \times 10^{- 6}N$
Explanation
(b) $F = qvB$ also kinetic energy $K = \frac{1}{2}mv^{2} \Rightarrow v = \sqrt{\frac{2K}{m}}$
$\therefore F = q\sqrt{\frac{2K}{m}}B$
$= 1.6 \times 10^{- 19}\sqrt{\frac{2 \times 200 \times 10^{6} \times 1.6 \times 10^{- 19}}{1.67 \times 10^{- 27}}} \times 5 = 1.6 \times 10^{- 10}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.