Question

The horizontal component of the earth's magnetic field at a place is $3.0\times {10}^{-4}\text{T}$ and the dip is ${53}^{\circ }$. A metal rod of length $25\text{cm}$ is placed in the north-south direction and is moved with a constant speed of $10{\text{cm s}}^{-1}$ towards the east. Calculate the emf induced in the rod.

• A. $50\mu \text{V}$
• B. $10\mu \text{V}$
• C. $30\mu \text{V}$
• D. $20\mu \text{V}$

Answer 1

The correct option is B $10\mu \text{V}$

The dip at a place is given by
$\tan \delta =\frac{B_V}{B_H}$

$\Rightarrow B_V=B_H\tan \delta =(3.0\times {10}^{-4})\tan {53}^0=4.0\times {10}^{-4}\text{T}$

The emf induced is,
$ϵ=vBl=0.1\times 4.0\times {10}^{-4}\times 0.25$
$=10\mu \text{V}$