Question:
Electromagnetic radiations are emitted from a \(15\,\text{W}\) point source. The peak value of the magnetic field at a distance of \(5\,\text{m}\) from the source is
Electromagnetic radiations are emitted from a \(15\,\text{W}\) point source. The peak value of the magnetic field at a distance of \(5\,\text{m}\) from the source is
Show Hint
For an isotropic source,
\[
I=\frac{P}{4\pi r^2}.
\]
For an electromagnetic wave,
\[
I=\frac{c}{2\mu_0}B_0^2
\]
and
\[
E_0=cB_0.
\]
These relations allow direct calculation of the peak electric or magnetic field from the power of the source.
Updated On: Jul 9, 2026
- \(4\times10^{-8}\,\text{T}\)
- \(2\times10^{-8}\,\text{T}\)
- \(2\times10^{-7}\,\text{T}\)
- \(4\times10^{-7}\,\text{T}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
The intensity of electromagnetic radiation at a distance \(r\) from an isotropic point source is
\[
I=\frac{P}{4\pi r^2}.
\]
Also,
\[
I=\frac{c}{2\mu_0}B_0^2,
\]
where
\[
B_0=\text{peak magnetic field}.
\]
Step 1: Calculate the intensity at \(5\,\text{m}\). Given, \[ P=15\,\text{W}, \qquad r=5\,\text{m}. \] Therefore, \[ I = \frac{15}{4\pi(5)^2}. \] \[ I = \frac{15}{100\pi}. \] \[ I \approx 4.77\times10^{-2}\,\text{W m}^{-2}. \]
Step 2: Use the relation between intensity and magnetic field. \[ I=\frac{c}{2\mu_0}B_0^2. \] Hence, \[ B_0 = \sqrt{\frac{2\mu_0 I}{c}}. \] Substituting \[ \mu_0=4\pi\times10^{-7}\,\text{H m}^{-1}, \qquad c=3\times10^8\,\text{m s}^{-1}, \] \[ B_0 = \sqrt{ \frac{ 2(4\pi\times10^{-7})(4.77\times10^{-2}) }{ 3\times10^8 } }. \] \[ B_0 \approx 2\times10^{-8}\,\text{T}. \]
Step 3: Write the final answer. \[ \boxed{B_0=2\times10^{-8}\,\text{T}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Calculate the intensity at \(5\,\text{m}\). Given, \[ P=15\,\text{W}, \qquad r=5\,\text{m}. \] Therefore, \[ I = \frac{15}{4\pi(5)^2}. \] \[ I = \frac{15}{100\pi}. \] \[ I \approx 4.77\times10^{-2}\,\text{W m}^{-2}. \]
Step 2: Use the relation between intensity and magnetic field. \[ I=\frac{c}{2\mu_0}B_0^2. \] Hence, \[ B_0 = \sqrt{\frac{2\mu_0 I}{c}}. \] Substituting \[ \mu_0=4\pi\times10^{-7}\,\text{H m}^{-1}, \qquad c=3\times10^8\,\text{m s}^{-1}, \] \[ B_0 = \sqrt{ \frac{ 2(4\pi\times10^{-7})(4.77\times10^{-2}) }{ 3\times10^8 } }. \] \[ B_0 \approx 2\times10^{-8}\,\text{T}. \]
Step 3: Write the final answer. \[ \boxed{B_0=2\times10^{-8}\,\text{T}} \] \[ \boxed{\text{Answer = (B)}} \]
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