There is a ring of radius $r$ having linear charge density $\lambda$ and rotating with a uniform angular velocity $\omega$. The magnitude of the magnetic field produced by this ring at its own centre would be
($\mu_0 = \text{permeability of air}$)
Magnetic Field Calculation for Rotating Ring
A rotating charged ring constitutes an electric current, which generates a magnetic field. We can calculate the magnetic field at the center by determining the equivalent current first.
Deriving the Magnetic Field Formula
Let the ring have radius $r$ and linear charge density $\lambda$. The total charge $Q$ on the ring is:
$Q = \lambda \times (2\pi r)$
The ring rotates with a constant angular velocity $\omega$. The time period $T$ for one full rotation is:
$T = \frac{2\pi}{\omega}$
The effective current $I$ generated by the rotating charge is charge passing per unit time:
$I = \frac{Q}{T} = \frac{\lambda \times (2\pi r)}{\frac{2\pi}{\omega}} = \lambda r \omega$
The magnetic field $B$ at the center of a circular current loop of radius $r$ is given by:
$B = \frac{\mu_0 I}{2r}$
Substitute the derived current $I$ into the magnetic field formula:
$B = \frac{\mu_0 (\lambda r \omega)}{2r}$
Simplifying the expression yields the magnetic field magnitude:
$B = \frac{\mu_0 \lambda \omega}{2}$
Final Magnetic Field Magnitude
The magnitude of the magnetic field produced by the rotating ring at its center is $\frac{\mu_0 \lambda \omega}{2}$.
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