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

A conducting ring of radius r is placed in a magnetic field B that varies with time. According to Faraday's Law of Electromagnetic Induction, the electromotive force (emf) induced in the ring is given by the rate of change of magnetic flux through the ring.

The magnetic flux Φ through the ring is Φ = B × A, where A is the area of the ring. Since the field is perpendicular to the plane of the ring, A = πr².

The rate of change of the magnetic flux is:

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

According to 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 given by the relation:

$$ε = E \cdot (2πr)$$

Substitute the value of ε:

$$-πr² \times x = E \cdot (2πr)$$

Solve for the electric field intensity E:

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

The magnitude of the electric field intensity, ignoring the negative sign (indicating direction), is:

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

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