Problem 150 - A Complete Resource Book in Physics for JEE Main

Question

In the diagram, there are four conducting plates \(A, B\), \(C\), and \(D\) placed parallel to each other at equal separation \(L\). If plate \(C\) starts moving towards plate \(B\) with velocity \(v\). Find the current (in mA) flowing in the wire connecting \(A\) and \(D\) (assume all other plates to be fixed). \(\left(\right.\) Given, \(\left.q_{2}=2 \mu \mathrm{C}, q_{3}=3 \mu \mathrm{C}, v=10 \mathrm{~m} / \mathrm{s}, L=0.05 \mathrm{~m}\right)\)

Step-by-Step Solution

Answer

To solve this problem, one needs to make use of Faraday's law of electromagnetic induction and the formula for electric field. After calculating the change in electric field using the provided data, one then calculates the induced emf and hence the current using the given formulas.

Step 1: Identify the relevant equations

The first step is to identify the relevant equations to solve the problem. Faraday's law of electromagnetic induction is the main equations used: \(E = -\frac{dΦ}{dt}\), where E is the electromotive force, Φ is the magnetic flux.

Step 2: Determine the change in electric field

In this scenario, when plate C moves towards plate B, the separation between them reduces but the charges remain the same. So, the electric field between the plates B and C changes. The electric field, which is given by \(E = \frac{q}{4πε0L}\), where q is the charge on the plate, ε0 is the permittivity of free space, and L is the separation.

Step 3: Find the induced emf

Using Faraday's law of electromagnetic induction, the induced electromotive force (emf) in the circuit would be \(E = -\frac{dΦ}{dt}\), where Φ = εE, E is the electric field, and ε is the permittivity of free space. Since the electric field E is changing due to the change in distance between the plates, dΦ/dt can be calculated as \(ε\frac{dE}{dt}\). This gives the rate of change in electric field, which when plugged in the Faraday's law equation will give the induced emf.

Step 4: Compute the current

The induced current in the wire connecting A and D can be determined by dividing the induced emf by the resistance (R) of the circuit. Since resistance is not mentioned in the problem, it can be assumed that the resistance is negligible. So, the current can be directly taken as \(I = \frac{E}{R}\), where E is the induced emf and R the resistance.

Key Concepts

Current in Circuit

When it comes to understanding the current flowing through a circuit, it's important to grasp how current behaves. Current, in its simplest form, is the flow of electric charge through a conductor. In this exercise's scenario, plates are conducting, implying they allow charges to move freely when connected by a wire.

As plate C moves toward plate B, the induced electric charge rearranges due to changing magnetic and electric fields. Given the principle of low-resistance conductors, we anticipate minimal resistance, allowing nearly full transfer of any induced electromotive force (EMF) to be observed as current.

Therefore, the induced EMF translates fully to current flow between plates A and D, when considering a negligible resistance in the circuit.

Electric Field Change

Understanding the change in the electric field is crucial here. Electric fields come into play between charged objects, such as the plates. Prior to movement, a certain electric field is established due to the charges on plates B and C.

When plate C moves, the distance between it and plate B decreases, altering the electric field in between. According to the formula \( E = \frac{q}{4\pi\epsilon_0 L} \), shortening the distance between plates increases the electric field strength because \( L \), the separation, is in the denominator.

Since \( q \) (the charge) remains constant, only changes in \( L \) affect the electric field. This shift is central to creating an induced EMF and current in our exercise.

Induced EMF Calculation

The calculation of induced EMF (Electromotive Force) relies on the fundamental principles of Faraday’s Law, which describes how a change in magnetic flux can form an EMF. In our scenario, although magnetic flux isn’t directly mentioned, the electric field changes contribute to a similar phenomenon.

Using Faraday's law of electromagnetic induction, we express:

• \( E = -\frac{d\Phi}{dt} \)
• where \( \Phi \) is related to the electric field as \( \Phi = \epsilon E \).

As the electric field \( E \) changes due to the distance \( L \) changing, we determine the rate of change \( \frac{dE}{dt} \), yielding the rate at which the field shifts over time. Plugging this change into Faraday's formula gives us the induced EMF. In simple terms, the swifter the separation between B and C alters, the greater the resultant EMF in the circuit.