Q. At a place the value of horizontal component of the earth's magnetic field $H$ is $3 \times 10^{-5}\, Wb / m ^{2}$. 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?
Chhattisgarh PMTChhattisgarh PMT 2005
Solution:
Induced potential difference in the rod.
$e=B v l$ or $e=H v l$
Given: $H=3 \times 10^{-5} \,Wb / m ^{2}$,
$v=50\, m / s$,
$l=2 \,m$
$\therefore e=3 \times 10^{-5} \times 50 \times 2$
$=3 \times 10^{-3}\, V =3 \,mV$
Now, by Right hand palm rule.
If $H$ is towards $N$ and rod is moving downward,
then the current will move form $B$ to $A$.
Hence, end $B$ will become positive.
