A conducting rod PQ of length 1 m is moving with a uniform speed v = 2m/sec in a uniform magnetic field B = 4 T

A capacitor C = 10 micro farad is connected as shown then find the charge the charge on plate A and Plate B

Ans qA = + 80 micro C

qB = -80 micro C

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To solve the problem of finding the charge on plates A and B of the capacitor connected to a conducting rod moving in a magnetic field, we can break it down into several logical steps. This involves using concepts from electromagnetism, specifically Faraday's law of electromagnetic induction and the relationship between charge, capacitance, and voltage.

Understanding the Setup

We have a conducting rod PQ of length 1 meter moving with a speed of 2 m/s in a magnetic field of 4 T (Tesla). The rod is part of a circuit that includes a capacitor with a capacitance of 10 microfarads (µF). The movement of the rod in the magnetic field induces an electromotive force (emf) across the rod, which in turn affects the charge on the capacitor plates.

Calculating the Induced EMF

According to Faraday's law, the induced emf (ε) in a circuit is given by the formula:

• ε = B * L * v

Where:

• B = magnetic field strength (4 T)
• L = length of the rod (1 m)
• v = speed of the rod (2 m/s)

Substituting the values:

• ε = 4 T * 1 m * 2 m/s = 8 V

Finding the Charge on the Capacitor

The charge (Q) on a capacitor is related to the voltage (V) across it and its capacitance (C) by the formula:

• Q = C * V

Here, we know the capacitance is 10 µF, which is equal to 10 x 10^-6 F, and the voltage across the capacitor is the induced emf we just calculated (8 V). Now we can find the charge:

• Q = 10 x 10^-6 F * 8 V = 80 x 10^-6 C = 80 µC

Charge Distribution on the Plates

In a capacitor, one plate accumulates positive charge while the other accumulates an equal amount of negative charge. Therefore, if plate A accumulates a charge of +80 µC, then plate B will have a charge of -80 µC. This is due to the principle of conservation of charge, which states that the total charge in an isolated system remains constant.

Final Results

Thus, the charges on the plates of the capacitor are:

• Charge on plate A (qA) = +80 µC
• Charge on plate B (qB) = -80 µC

In summary, the movement of the conducting rod in the magnetic field induces an emf, which leads to a charge accumulation on the capacitor plates, resulting in +80 µC on plate A and -80 µC on plate B. This example illustrates the fundamental principles of electromagnetism and capacitance in action.