Question

A rod $PQ$ is connected to the capacitor plates. The rod is placed in a magnetic field $\left(B\right)$ directed downward perpendicular to the plane of the paper. If the rod is pulled out of magnetic field with velocity $\overrightarrow{V}$ as shown in fig. Then, identify the correct statement :

• A. Plate $M$ will be positively charged
• B. Plate $N$ will be positively charged
• C. both plates will be similarly charged
• D. No charge will be collected on either plates

Answer 1

The correct option is A Plate $M$ will be positively charged

$\Rightarrow$The sign of Motional emf $\left(\epsilon \right)$ produced across the ends of a rod is given by:
$\overrightarrow{V}\times \overrightarrow{B}$, so using right hand rule for cross product, the direction of thumb will point towards $+ve$ terminal.
$\Rightarrow$direction of $\overrightarrow{V}$ is $\hat{i}$, direction of $\overrightarrow{B}$ is $-\hat{k}$
$\therefore \hat{i}\times -\hat{k}=\hat{j}$ i.e along $+vey-$ direction for $(+ve)$ terminal of motional emf.
$\Rightarrow$End $P$ will be connected to $+ve$ terminal of emf $\epsilon$, hence plate $M$ of capacitor will become positively charged.