Question

The efficiency of a bulb of \(1000~\text{W}\) is \(6%\). The peak value of the magnetic field produced by the electromagnetic radiation from the bulb at a distance of \(1~\text{m}\) is: \[ \left( \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9~\text{Nm}^2\text{C}^{-2} \right) \]

• A. \(20 \times 10^{-8}~\text{T}\)
• B. \(15 \times 10^{-8}~\text{T}\)
• C. \(10 \times 10^{-8}~\text{T}\)
• D. \(5 \times 10^{-8}~\text{T}\)

Hint:

Use \(B_0 = \frac{E_0}{c}\), where \(E_0 = \sqrt{ \frac{2I}{c\varepsilon_0} }\); and \(I = \frac{P}{4\pi r^2}\)

Answer 1

The Correct Option is A

Useful power emitted as radiation = \(6% \text{ of } 1000~\text{W} = 60~\text{W}\) The intensity \(I\) at \(r = 1~\text{m}\): \[ I = \frac{P}{4\pi r^2} = \frac{60}{4\pi (1)^2} = \frac{60}{4\pi} \] Peak electric field: \[ E_0 = \sqrt{ \frac{2I}{c\varepsilon_0} } \quad \text{and} \quad B_0 = \frac{E_0}{c} \Rightarrow B_0 = \sqrt{ \frac{2I}{c^3 \varepsilon_0} } \] Using simplified relation for \(B_0\): \[ B_0 = \sqrt{ \frac{2 \times 60}{(3 \times 10^8)^3 \cdot 8.854 \times 10^{-12}} } \approx 20 \times 10^{-8}~\text{T} \]

Answer 2

Step 1: Understand the problem and given data
- Power of bulb, \(P = 1000~\text{W}\)
- Efficiency, \(\eta = 6\% = 0.06\)
- Distance from bulb, \(r = 1~\text{m}\)
- \(\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9~\text{Nm}^2\text{C}^{-2}\) (Given, but not needed directly for this calculation)

Step 2: Calculate the power radiated as electromagnetic waves
Power radiated, \(P_{\text{rad}} = \eta \times P = 0.06 \times 1000 = 60~\text{W}\)

Step 3: Use the intensity formula at distance \(r\)
Intensity, \(I = \frac{P_{\text{rad}}}{4 \pi r^2} = \frac{60}{4 \pi (1)^2} = \frac{60}{4 \pi} = \frac{60}{12.566} \approx 4.77~\text{W/m}^2\)

Step 4: Relate intensity to electric and magnetic field amplitudes
Intensity is related to the peak electric field \(E_0\) and magnetic field \(B_0\) by:
\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 = \frac{1}{2} \frac{c B_0^2}{\mu_0} \]
Since \(c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\), the formula for magnetic field amplitude is:
\[ B_0 = \sqrt{\frac{2 \mu_0 I}{c}} \]

Step 5: Use constants values
- Speed of light, \(c = 3 \times 10^8~\text{m/s}\)
- Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}~\text{T·m/A}\)

Step 6: Calculate \(B_0\)
\[ B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 4.77}{3 \times 10^8}} = \sqrt{\frac{1.2 \times 10^{-5}}{3 \times 10^8}} \approx \sqrt{4 \times 10^{-14}} = 2 \times 10^{-7}~\text{T} \]

Step 7: Final answer
The peak magnetic field is:
\[ B_0 = 20 \times 10^{-8}~\text{T} \]