Question

A non-conducting ring of radius R having uniformly distributed charge Q starts rotating about x-x' axis passing through diameter with an angular acceleration $\alpha$ as shown in the figure. Another small conducting ring having radius a (a << R) is kept fixed at the centre of bigger ring is such a way that axis xx' is passing through its centre and perpendicular to its plane. If the resistance of small ring is r = 1$\Omega$, find the induced current in it in ampere.

(Given $q = \frac{16 \times 10^2}{\mu_0}C$, R = 1 m, a = 0.1 m, $\alpha = 8$ rad/s$^2$)

Answer 1

16 A

Answer 2

Solution

1. Magnetic Field Calculation:

   Consider a small element $dq=\frac{Q}{2\pi}d\phi$ on the big ring of radius $R$ (which carries a uniformly distributed charge $Q$).

   When the ring rotates about the $x$–axis (a diameter) with angular speed $\omega$ (with $\omega=\alpha t$ from rest), each charge element has velocity $v=\omega R$ (with appropriate direction).

   Using the Biot–Savart law, the contribution of each element to the $x$–component of the magnetic field at the centre is

   $$dB_x=\frac{\mu_0}{4\pi}\frac{dq\,(r_yv_z-r_zv_y)}{R^3}.$$

   A careful integration over the ring (taking into account that in our chosen geometry the only surviving component is along the $x$–axis) gives

   $$B_x=\frac{\mu_0\,Q\,\omega}{8\pi R}.$$

2. Substitute the Given Charge $Q$:

   The problem gives

   $$Q=\frac{16\times10^2}{\mu_0}=\frac{1600}{\mu_0}\quad \text{(in Coulombs)}.$$

   Substitute in the expression for $B_x$:

   $$B_x=\frac{\mu_0\,(1600/\mu_0)\,\omega}{8\pi R}=\frac{1600\,\omega}{8\pi R}=\frac{200\,\omega}{\pi R}.$$

3. Flux Through the Small Ring:

   The small ring (radius $a$) is placed at the centre with its plane perpendicular to the $x$–axis. Its area is $A=\pi a^2$.

   Thus the magnetic flux is

   $$\Phi = B_x\,A=\frac{200\,\omega}{\pi R}\cdot\pi a^2=\frac{200\,\omega\,a^2}{R}.$$

4. Induced Emf and Current:

   By Faraday’s law, the induced emf is

   $$\mathcal{E}=-\frac{d\Phi}{dt}=-\frac{200\,a^2}{R}\frac{d\omega}{dt}.$$

   Since $\frac{d\omega}{dt}=\alpha$, we obtain

   $$\mathcal{E}=-\frac{200\,a^2\alpha}{R}.$$

   Taking the magnitude (and using the resistance $r=1\,\Omega$) the induced current is

   $$I=\frac{|\mathcal{E}|}{r}=\frac{200\,a^2\alpha}{R}.$$

   Now, substituting the numerical values $a=0.1\,\text{m},\;\alpha=8\,\text{rad/s}^2,\;R=1\,\text{m}$:

   $$I=\frac{200\times(0.1)^2\times8}{1}=\frac{200\times0.01\times8}{1}=16\,\text{A}.$$

Final Answer:

The induced current in the small ring is 16 A.