Question

Assume Earth's surface is a conductor with a uniform surface charge density $\sigma$. It rotates about it's axis with angular velocity $\omega$. Suppose the magnetic field due to Sun at Earth at some instant is a uniform field $B$ pointing along Earth's axis. Then the emf developed between the pole and equator of earth due to this field is
($R_e=$ radius of earth)

• A. $\frac{1}{2}B\omega R_e^2$
• B. $B\omega R_e^2$
• C. $\frac{3}{2}B\omega R_e^2$
• D. zero

Answer 1

The correct option is A $\frac{1}{2}B\omega R_e^2$

The equator can be seen as a conducting ring of radius $R_e$ revolving with angular velocity $\omega$ in a perpendicular magnetic field B.
$\therefore$ Potential difference. across its center and periphery
$=\frac{B\omega R_e^2}{2}$
Potential at pole $=$ potential of the axis of earth i.e. potential at point O
$\therefore V_{equator}-V_{pole}=\frac{B\omega R{{}_e}^2}{2}.$