A uniformly charged ring of radius `0.1 m` rotates at a frequency of `10^(4) rps` about its axis.Find the ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance `0.2 m` from the centre.(Use speed of light `c=3xx10^(8) m//s,pi^(2)=10`)

Explanation

To solve the problem, we need to find the ratio of the energy density of the electric field (\(U_E\)) to the energy density of the magnetic field (\(U_B\)) at a point on the axis of a uniformly charged rotating ring.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the ring, \(a = 0.1 \, \text{m}\)
- Frequency of rotation, \(f = 10^4 \, \text{rps}\)
- Distance from the center on the axis, \(x = 0.2 \, \text{m}\)
- Speed of light, \(c = 3 \times 10^8 \, \text{m/s}\)
- \(\pi^2 = 10\)

2. **Calculate Angular Frequency (\(\omega\)):**
\[ \omega = 2\pi f = 2\pi \times 10^4 \, \text{rad/s} \]

3. **Electric Field (\(E\)) at a point on the axis:**
The electric field at a distance \(x\) from the center of the ring is given by:
\[ E = \frac{kQx^3}{(x^2 + a^2)^{3/2}} \]
where \(k = \frac{1}{4\pi \epsilon_0}\).

4. **Magnetic Field (\(B\)) at a point on the axis:**
The magnetic field due to a rotating charged ring can be expressed as:
\[ B = \frac{\mu_0 I}{2a} \]
where \(I = Q\omega\) (current due to rotation).

5. **Energy Densities:**
The energy density of the electric field is given by:
\[ U_E = \frac{1}{2} \epsilon_0 E^2 \]
The energy density of the magnetic field is given by:
\[ U_B = \frac{1}{2} \frac{B^2}{\mu_0} \]

6. **Ratio of Energy Densities:**
The ratio of the energy densities can be expressed as:
\[ \frac{U_E}{U_B} = \frac{\epsilon_0 E^2}{\frac{B^2}{\mu_0}} = \frac{\epsilon_0 E^2 \mu_0}{B^2} \]

7. **Substituting Values:**
Substitute \(E\) and \(B\) into the ratio:
\[ \frac{U_E}{U_B} = \frac{\epsilon_0 \left(\frac{kQx^3}{(x^2 + a^2)^{3/2}}\right)^2 \mu_0}{\left(\frac{\mu_0 I}{2a}\right)^2} \]

8. **Simplifying the Ratio:**
After simplification, we find that the terms involving \(Q\) cancel out, leading to a simplified expression for the ratio.

9. **Final Calculation:**
Substitute the known values into the final expression to find the numerical value of the ratio.

10. **Conclusion:**
The final ratio of the energy density of the electric field to the energy density of the magnetic field is:
\[ \frac{U_E}{U_B} = 9 \times 10^9 : 1 \]