Question

A metallic block carrying current $i$ is subjected to a uniform magnetic field $\overrightarrow{B}$ as shown in figure. The moving charges in the metal experience a force $F$ given by _______ , which results in the lowering of the electric potential of the face_______. Assume the speed of the charged particle to be $v$, then select the option to fill in the blanks.

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

Answer 1

The correct option is C $evB\hat{k},\text{ABDC}$

The moving charge carrier in the metallic block are free electrons. So, if the direction of current is along $+vex-$ axis, then the electrons are moving along $-vex-$ axis.


Applying the equation of magnetic force acting on the moving charged particle in a magnetic field,

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

$\overrightarrow{F}=-e(-v\hat{i}\times B\hat{j}]=evB\hat{k}$

So, the force on electrons is directed towards face $\text{ABDC}$.

Thus, the electrons will tend to accumulate at that face, and it will become negatively charged, and to maintain the neutrality of the metal, the opposite face will bear an equal amount of positive charge.

Therefore, electric potential of face $\text{ABDC}$ will get lowered.

Why this question ? Tip: The face $\text{ABDC}$ will become negatively charged due to accumulation of electrons, thus, an electric field will develop from face $\text{EFHG}$ to face $\text{ABDC}$, due to produced potential difference between these faces.