Q. At a place the value of horizontal component of the earth’s magnetic field $H$ is $3 \times 10^{- 5}weber/m^{2}$. A metallic rod $AB$ of length $2\ m$ placed in east-west direction, having the end $A$ towards east, falls vertically downward with a constant velocity of $50\ m/s$. Which end of the rod becomes positively charged and what is the value of induced potential difference between the two ends
A) End $A,\ 3 \times 10^{- 3}mV$
B) End $A,3\ mV$
C) End $B,\ 3 \times 10^{- 3}\ mV$
D) End $B,\ 3\ mV$
Explanation
(b) Induced potential difference between two ends $= Blv = B_{H}lv$
$= 3 \times 10^{- 5} \times 2 \times 50 = 30 \times 10^{- 3}volt = 3\ millivolt$
By Fleming’s right hand rule, end $A$ becomes positively charged
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.