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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Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
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First Configuration
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