Question

A long metal bar of $30\text{cm}$ length is aligned along a north-south line and is moving eastward, at a speed of $10{\text{ms}}^{-1}$. A uniform magnetic field of $4.0\text{T}$ points vertically downwards. If the south end of the bar has a potential of $0\text{V}$, the induced potential at the north end of the bar is :

• A. $+12\text{V}$
• B. $-12\text{V}$
• C. $0\text{V}$
• D. $+10\text{V}$

Answer 1

The correct option is A $+12\text{V}$


The induced emf,

$\epsilon =vBl=10\times 4\times 0.3=12\text{V}$

From right-hand rule, the northern end will be at higher potential.


As the southern end is at $0\text{V}$, the northern end must be at $+12\text{V}$.

Hence, option $\left(A\right)$ is the correct answer.