MHT CET · Physics · Magnetic Effects of Current
A thin ring of radius ' \(R\) ' carries a uniformly distributed charge. The ring rotates at constant speed ' \(N\) ' r.p.s. about its axis perpendicular to the plane. If ' \(B\) ' is the magnetic field at the centre, the charge on the ring is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 \mathrm{~N}}{2 \mathrm{RB}}\)
- B \(\frac{\mathrm{RB}}{2 \mu_0^{\prime} \mathrm{N}}\)
- C \(\frac{\mu_0 \mathrm{~N}}{\mathrm{RB}}\)
- D \(\frac{2 R B}{\mu_0 \mathrm{~N}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 R B}{\mu_0 \mathrm{~N}}\)
Step-by-step Solution
Detailed explanation
A thin uniformly charged rotating ring acts like a current carrying coil.
Magnetic field at the centre of current carrying coil,
\(B=\frac{\mu_0 I}{2 R}\)
Current is given by, \(\mathrm{I}=\mathrm{qN} \quad\) where N is revolutions per second.
\(\begin{array}{ll}
\therefore & B=\frac{\mu_0 q N}{2 R} \\
\therefore & q=\frac{2 R B}{\mu_0 N}
\end{array}\)
...[From(i)]
Magnetic field at the centre of current carrying coil,
\(B=\frac{\mu_0 I}{2 R}\)
Current is given by, \(\mathrm{I}=\mathrm{qN} \quad\) where N is revolutions per second.
\(\begin{array}{ll}
\therefore & B=\frac{\mu_0 q N}{2 R} \\
\therefore & q=\frac{2 R B}{\mu_0 N}
\end{array}\)
...[From(i)]
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