Question

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

• A. \(1.19 × 10^{–8}\;T\)
• B. \(1.71 × 10^{–8}\;T\)
• C. \(0.84 × 10^{–8}\;T\)
• D. \(3.36 × 10^{–8}\;T\)

Answer 1

The Correct Option is B

To determine the peak magnetic field at a 4 m distance from a bulb rated at 200 W operating as a point source with 3.5% efficiency, follow the steps below:

1. Calculate the Power Radiated as Electromagnetic Waves:

   The bulb has an efficiency of 3.5%, which means that only 3.5% of the total power is converted into electromagnetic radiation.

   Total Power output of bulb \( = 200 \, \text{W} \)

   Efficiency \( = 3.5\% \)

   Power radiated as electromagnetic waves is given by:

   P_{\text{radiated}} = 0.035 \times 200 = 7 \, \text{W}

2. Understanding Intensity:

   Intensity (I) is the power per unit area and is given by the formula:

   I = \frac{P_{\text{radiated}}}{4\pi d^2} = \frac{7}{4\pi (4)^2} = \frac{7}{64\pi} \, \text{W/m}^2

3. Relate Intensity to Peak Magnetic Field:

   The intensity (I) and peak magnetic field (B_0) in an electromagnetic wave are related as:

   I = \frac{1}{2}\cdot c\cdot\mu_0\cdot B_0^2, where c = 3 \times 10^8 \, \text{m/s} and \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}.

   Rearranging for B_0, we get:

   B_0 = \sqrt{\frac{2I}{c\mu_0}}

4. Calculate Peak Magnetic Field:

   Substitute the intensity (I) calculated earlier:

   B_0 = \sqrt{\frac{2 \times \frac{7}{64\pi}}{3 \times 10^8 \times 4\pi \times 10^{-7}}}

   = \sqrt{\frac{7}{384 \times 10^{-7}}}

   = 1.71 \times 10^{-8} \, \text{T}

This confirms that the correct answer is 1.71 \times 10^{-8}\;T.