Q. A $1\mu A$ beam of protons with a cross-sectional area of $0.5\ sq.\ mm$ is moving with a velocity of $3 \times 10^{4}ms^{- 1}.$ Then charge density of beam is
A) $6.6 \times 10^{- 4}C/m^{3}$
B) $6.6 \times 10^{- 5}C/m^{3}$
C) $6.6 \times 10^{- 6}C/m^{3}$
D) None of these
Explanation
(b) For one second, distance = Velocity $= 3 \times 10^{4}m/sec$ and $Q = i \times 1 = 10^{- 6}C.$ Charge density $= \frac{Charge}{Volume}$
$= \frac{10^{- 6}}{3 \times 10^{4} \times 0.5 \times 10^{- 6}} = 6.6 \times 10^{- 5}C/m^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.