Question

Figure shows a circular area of radius R where a uniform magnetic field $\vec{B}$ is going into the plane of paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r)?

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is D

 To solve this problem, we need to understand the concept of electromagnetic induction and the relationship between a changing magnetic field and the induced electric field. According to Faraday's law of electromagnetic induction, a changing magnetic field through a loop induces an electromotive force (EMF) in the loop, which can be related to an induced electric field \(E(r)\).

The given scenario describes a circular area with a uniform magnetic field \(\vec{B}\) that is increasing in magnitude at a constant rate. The magnetic field is directed into the plane of the paper.

Let's analyze step-by-step how the induced electric field varies with the distance \(r\) from the center of the circular area:

1. According to Faraday's law, the induced EMF around a closed loop is equal to the rate of change of magnetic flux through the loop. Mathematically, it is given by: \(\text{EMF} = -\frac{d\Phi}{dt}\).
2. For a circular loop of radius \(r\) within the circular area of radius \(R\) (where \(r \leq R\)), the magnetic flux \(\Phi\) is: \(\Phi = \pi r^2 B\).
3. As the magnetic field \(B\) is increasing at a constant rate, we denote this rate as \(\frac{dB}{dt}\). Therefore, the rate of change of flux is: \(\frac{d\Phi}{dt} = \pi r^2 \frac{dB}{dt}\).
4. The induced EMF around the loop is related to the electric field by \(\text{EMF} = 2\pi r E(r)\). Equating this to the rate of change of flux, we get: \(2\pi r E(r) = \pi r^2 \frac{dB}{dt}\).
5. Solving for \(E(r)\) gives: \(E(r) = \frac{r}{2} \frac{dB}{dt}\) for \(r \leq R\).
6. For \(r > R\), the magnetic flux through the loop is constant (since it is outside the region of changing magnetic field), so \(E(r) = 0\).

From these steps, we conclude that the induced electric field increases linearly with \(r\) inside the circular area and is zero outside. Thus, the graph of the induced electric field \(E(r)\) with respect to \(r\) should show a linear increase from 0 to R and then remain zero for \(r > R\).

Hence, the correct graph representing \(E(r)\) is shown in the above image.