Question

A conducting rod PQ of length $L = 1.0m$ is moving with a uniform speed $v = 2.0m/s$ in a uniform magnetic field $B = 4.0T$ directed into the plane of the paper. A capacity $C = 10\mu F$ is connected as shown in fig., then

A. ${q_A} = + 80\mu C$ and ${q_B} = - 80\mu C$
B. ${q_A} = - 80\mu C$ and${q_B} = + 80\mu C$
C. ${q_A} = 0 = {q_B}$
D. Charge stored in the capacitor increases exponentially with time.

Answer 1

Hint: The potential difference is calculated by multiplying magnetic field, length of the rod and velocity of the rod. Then the charge is obtained by the product of potential difference and capacitance and one side of the capacitor will develop positive charge and the other would develop negative charge.

Complete step by step solution:
An emf induced by motion relative to a magnetic field is called a motional emf. Let us assume that the conductor is moving in a direction perpendicular to the magnetic field with constant velocity under the influence of some external agent or force. The electrons in the conductor i.e, rod experiences a force which is given by the formula $F = - e\left( {v \times B} \right)$.

Under the influence of this force, the electrons move to the lower end of the conductor and accumulate there and a positive charge is created at the upper end. Due to this difference, an electric field is created into the conductor (rod).These accumulation of charge continues until downward magnetic force is balanced by the upward electric force i.e., in equilibrium $eE = evB$ where e is the charge on an electron, E is the electric field, v is the velocity of the rod and B is the magnetic field. We get $E = vB$ .

The electric field produced in the rod is related to the potential difference (V) across the ends of the conductor i.e., rod $V = EL$ where L is the length of the rod. On substituting the value of E in this, we get $V = BvL$ .
$V = 4 \times 2 \times 1$
$\Rightarrow V = 8V$
A capacitor AB is connected to this and let the charge induced in it be Q.
$Q = CV$ where C is the capacitance of the capacitor
$\Rightarrow Q = 10 \times 8$
$\therefore Q = 80\mu C$
At A the charge induced is $ - 80\mu F$ and at B charge induced is $ + 80\mu F$ because the current flows in an anticlockwise direction.

Therefore, option B is correct.

Note: Remember the formula of emf or potential difference that is generated due to the moving conductor and the formula of charge developed in the capacitor due to this emf. The charge developed at A is negative and charge developed at B is positive due to the movement of electrons from one side to another side in a conductor as the opposite charge is induced in a capacitor as compared to the moving conductor.