Question

A conducting rod $\text{PQ}$ of length $l=1.0\text{m}$ is moving with a uniform speed $v=20\text{m/s}$ in a uniform magnetic field $B=4.0\text{T}$ directed into the paper. A capacitor of capacity $C=10\mu \text{F}$ is connected as shown in figure. Then

• A. $q_A=+800\mu C\text{and}{q}_B=-800\mu C$
• B. $q_A=-800\mu C\text{and}{q}_B=+800\mu C$
• C. $q_A=q_B=0$
• D. $\text{The charged stored in the capacitor}\text{increases exponentially with time.}$

Answer 1

The correct option is A $q_A=+800\mu C\text{and}{q}_B=-800\mu C$

Emf induced across the rod or capacitor is,

$E=Bvl$

$\Rightarrow E=4\times 20\times 1=80\text{V}$

$\therefore$ Charge on the capacitor is,

$q=CE$

$=10\times {10}^{-6}\times 80=800\mu \text{C}$

According to the Lenz’s law, the direction of the current induced in the rod by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field $B$ which produced it. The direction of this current flow is given by Fleming’s right hand rule.

According to Fleming’s right-hand rule, induced current flows from $Q$ to $P$. Hence, $P$ is at higher potential and $Q$ is at lower potential. Therefore, $A$ is positively charged and $B$ is negatively charged. Hence,

$\therefore$ Charge on plate $\text{A}=800\mu \text{C}$

Charge on plate $\text{B}=-800\mu \text{C}$


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

Why this question? This question challenges your ability to calculate the direction of induced emf in the circuit.