Question

A conducting ring of radius r is placed in a varying magnetic field perpendicular to the plane of the ring. If the rate at which the magnetic field varies is x, the electric field intensity at any point of the ring is________.
Fill in the blank with the correct answer from the options given below.

• A. rx/2
• B. rx
• C. 2rx
• D. 4r/x

Answer 1

The Correct Option is A

Let's analyze the induced electric field in the conducting ring due to the varying magnetic field.

1. Faraday's Law of Electromagnetic Induction:

The induced emf (ε) in the ring is given by:

ε = -dΦ/dt

Where:

• Φ is the magnetic flux
• t is time

2. Magnetic Flux (Φ):

Φ = BA

Where:

• B is the magnetic field
• A is the area of the ring (A = πr²)

Therefore:

Φ = Bπr²

3. Induced EMF (ε):

ε = -d(Bπr²)/dt

ε = -πr²(dB/dt)

We are given that the rate at which the magnetic field varies is x, so dB/dt = x:

ε = -πr²x

4. Induced Electric Field (E):

The induced emf is also related to the electric field (E) by:

ε = ∫E·dl

For the circular ring, the line integral becomes:

ε = E(2πr)

Equating the two expressions for ε:

-πr²x = E(2πr)

E = -πr²x / (2πr)

E = -rx/2

The magnitude of the electric field intensity is rx/2.

Therefore, the electric field intensity at any point of the ring is rx/2.

The correct answer is:

Option 1: rx/2