Question

Assume a bulb of efficiency of 2.5% as a point source. The peak value of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

• A. 2.5 V$m^{-1}$, $3.6\times {10}^{-8}$T
• B. 4.2 V$m^{-1}$, $2.8\times {10}^{-8}$T
• C. 4.08 V$m^{-1}$, $1.36\times {10}^{-8}$T
• D. 3.6 V$m^{-1}$, $4.2\times {10}^{-8}$T

Answer 1

The correct option is C 4.08 V$m^{-1}$, $1.36\times {10}^{-8}$T

$I=\frac{power}{area}=\frac{100\times \left(2.5/10\right)}{4\pi (3{)}^2}=\frac{2.5}{36\pi }W{m}^{-2}$

Half of this intensity (I) belongs to electric field and half of that of magnbetic field. Therefore,

$\frac{I}{2}=\frac{1}{4}{ϵ}_0{E}_0^{2}C$

or $E_0=\sqrt{\frac{2I}{{ϵ}_{0}C}}$

$=\sqrt{\frac{2\times \left(2.5/36\pi \right)}{\left(\frac{1}{4\pi \times 9\times {10}^9}\right)\times (3\times {10}^8)}}$

$=4.08V{m}^{-1}$

$B_0=\frac{E_0}{c}=\frac{4.08}{3\times {10}^8}=1.36\times {10}^{-8}T$