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Question:

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is

μ₀q/(2πfR)

μ₀qf/(2πR)

μ₀qf/(2R)

μ₀q/(2fR)

Solution:

B = μ₀I/(2R)
where I is the current. The charge q rotates with frequency f Hz. Therefore, the time period is T = 1/f.
The current I is given by the charge flowing per unit time, I = q/T = qf.
Substituting this into the equation for B, we get
B = μ₀qf/(2R)