Question

A metallic block carrying current $I$ is subjected to a uniform magnetic induction $\overrightarrow{B}$ as shown in the figure. Find the force experience $\overrightarrow{F}$ by the moving charges. Also find the face, whose potential is lowered result of this force. Assume the speed of the charge carriers to be $v$.

• A. $evB\hat{k},\text{ABCD}$
• B. $evB\hat{k},\text{EFGH}$
• C. $-evB\hat{k},\text{ABCD}$
• D. $-evB\hat{k},\text{EFGH}$

Answer 1

The correct option is A $evB\hat{k},\text{ABCD}$

Here, the charge carriers are electrons, so for current flowing along $+x$-direction, the electron will be moving towards $-x$-axis.

(i.e), $\overrightarrow{v}=-v\hat{i}$ and $B=B\hat{j}$

Now, magnetic force on the charge carriers is,

$\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})$

$=-e[-v\hat{i}\times B\hat{j}]$

$\therefore \overrightarrow{F_B}=evB\hat{k}$

This force will push the electrons towards the face $\text{ABCD}$, lowering its potential.

Hence, the face $\text{ABCD}$ will be at lower potential.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.

Why this question ? To check the understanding of Hall effect.