Question

Assume Earth’s surface is a conductor with a surface charge density \[\sigma \]. It rotates about its axis with angular velocity $\omega $. Suppose the magnetic field due to the Sun at Earth at some instant is a uniform field B pointing along Earth’s axis. Then the emf developed between the pole and the equator of the earth due to this field is (${R_e}$=radius of the earth)
A) $\dfrac{1}{2}B\omega {R_e}^2$
B) $B\omega {R_e}^2$
C) $\dfrac{3}{2}B\omega {R_e}^2$
D) Zero

Answer 1

Hint: The induced Emf has to be calculated. The induced Emf due to a magnetic field calculated at different points of Earth will be different. This will solve your current problem.

Formula Used:
The Emf induced due to the magnetic field is
$Emf = \dfrac{{BA}}{t}$
Where $B$ is the Magnetic field, $A$ is the Area, and $t$ is the change in time.

Complete step by step answer:
When a conductor is passed inside a magnetic field, an EMF is generated and the reaction caused due to the magnetic field induced in the conductor with the original magnetic field causes deflection.
Using Faraday's Law we have,
Induced EMF is given as the rate at which magnetic flux changes in time.
Induced EMF = (change in Magnetic Flux Density x Area)/change in Time.
Which can also be written as,
$Emf = \dfrac{{BA}}{t}$
$A = {R^2}\Delta x$
Where $R$ is the radius of the conductor and $\Delta x$ is the change in distance.
$\Rightarrow Emf = \dfrac{{B{R^2}\Delta x}}{t} \\$
$\Rightarrow Emf = B{R^2}\dfrac{{\Delta x}}{t} \\$
$\Rightarrow Emf = B{R^2}\omega \\ $
where $\omega $= the angular velocity of Earth.
Now, the Emf at the equator, considering the equator as a ring with a radius and an angular velocity, can be written as
\[Em{f_E} = B{R_e}^2\omega \]
where ${R_e}$= radius of the Earth, and $\omega $= angular velocity of Earth.
Also, the Emf induced at the pole can be given by,
$\Rightarrow Emf = B{R_e}^2\omega \\$
$\Rightarrow Emf = B \times 0 \times \omega \\$
$\Rightarrow Em{f_P} = 0 \\$
Thus, the resultant Emf induced due to the magnetic field of Earth will be,

$Em{f_T} = \dfrac{{Em{f_E} - Em{f_P}}}{2}$
Putting in the values we get,
$\Rightarrow Em{f_T} = \dfrac{{Em{f_E} - Em{f_P}}}{2} \\$
$\Rightarrow Em{f_T} = \dfrac{{B{R_e}^2\omega - 0}}{2} \\$
$\Rightarrow Em{f_T} = \dfrac{1}{2}B{R_e}^2\omega \\ $

Thus, the emf induced between the pole and the equator of earth is $\dfrac{1}{2}B{R_e}^2\omega $.

Hence, option (A) is the correct answer.

Note:
When we are solving a problem like this, we have to first visualize the object to get a clear picture. Then we have to calculate the parameter at the individual points. Subsequently by subtracting we get the final answer.