Q101 / 150 · MHT CET 2021 (23 Sep Shift 2) · Physics

A thin ring of radius ' $R$ ' meter has charge ' $q$ ' coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolution/s. The value of magnetic induction in $\mathrm{Wb} \mathrm{m}{ }^{-2}$ at the center of the ring is ( $\mu_0=$ Permeability of free space)

  • A
    $\frac{\mu_0 q \mathrm{f}}{2 \pi R}$
  • B
    $\frac{\mu_0 q}{2 \pi R}$
  • Correct answer
    C
    $\frac{\mu_0 \mathrm{qf}}{2 \mathrm{R}}$
  • D
    $\frac{\mu_0 \mathrm{q}}{2 \pi \mathrm{fR}}$

Solution

For circular coil of radius $\mathrm{R}$ carrying a current $\mathrm{I}$, the magnetic field at the center is given by $$ B=\frac{\mu_0 I}{2 R} $$ Here $\mathrm{I}=\mathrm{qf}$ $$ \therefore \mathrm{B}=\frac{\mu_0 \mathrm{qf}}{2 \mathrm{R}} $$

Hence, the correct answer is option: 'C'.

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