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.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( rx/2 \)
A changing magnetic field produces an induced electric field whose lines are circles. By Faraday's law, the line integral of this field round the ring equals the rate of change of flux through it.
\( \displaystyle \oint \vec{E}\cdot d\vec{l} = \frac{d\phi}{dt} = \pi r^{2}\,\frac{dB}{dt} \)
By symmetry \( E \) has the same magnitude at every point of the ring and is along the ring, so the left side is \( E \times 2\pi r \).
\( \displaystyle E\,(2\pi r) = \pi r^{2}\,x \)
\( \displaystyle E = \frac{\pi r^{2} x}{2\pi r} = \frac{rx}{2} \)