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. \(\frac {rx}{2}\)
• B. \(rx\)
• C. \(2rx\)
• D. \(\frac {4r}{x}\)

Answer 1

The Correct Option is A

An r-radius conducting ring is subjected to a time-varying magnetic field B. Faraday's Law of Electromagnetic Induction dictates that the induced electromotive force (emf) in the ring equals the negative rate of change of magnetic flux through it.

Magnetic flux Φ through the ring is defined as Φ = B × A, where A is the ring's area. Given the field is perpendicular to the ring's plane, A = πr².

The rate of change of magnetic flux is calculated as:

$$\frac{dΦ}{dt} = A \frac{dB}{dt} = πr² \times x$$

By Faraday's Law, the induced emf (ε) is:

$$ε = -\frac{dΦ}{dt} = -πr² \times x$$

For a circular loop, the relationship between emf and electric field intensity (E) is:

$$ε = E \cdot (2πr)$$

Substituting the value of ε yields:

$$-πr² \times x = E \cdot (2πr)$$

Solving for the electric field intensity E:

$$E = \frac{-πr² \times x}{2πr} = \frac{-rx}{2}$$

The magnitude of the electric field intensity, disregarding the directional negative sign, is:

$$E = \frac{rx}{2}$$

Consequently, the electric field intensity at any point on the ring is \(\frac {rx}{2}\).