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Question: Assume a bulb of efficiency 2.5% as a point source. The peak values of electric and magnetic fields ...

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

A

2.5Vm−1,3.6×10−8T2.5Vm^{- 1,}3.6 \times 10^{- 8}T

B

4.2Vm−1,2.8×10−8T4.2Vm^{- 1,}2.8 \times 10^{- 8}T

C

4.08Vm−1,1.36×10−8T4.08Vm^{- 1,}1.36 \times 10^{- 8}T

D

3.6Vm−1,4.2×10−8T3.6Vm^{- 1,}4.2 \times 10^{- 8}T

Answer

4.08Vm−1,1.36×10−8T4.08Vm^{- 1,}1.36 \times 10^{- 8}T

Explanation

Solution

: Here intensity, I=powerareaI = \frac{power}{area}

=100×2.54π(3)2×100=2.536πWm−2= \frac{100 \times 2.5}{4\pi(3)^{2} \times 100} = \frac{2.5}{36\pi}Wm^{- 2}

Half of this intensity belongs to electric field and half of that to magnetic field.

∴I2=14ε0E02c\therefore\frac{I}{2} = \frac{1}{4}\varepsilon_{0}E_{0}^{2}c or

E0=2Iε0c=2×2.536π14π×9×109×3×108=4.08Vm−1∴B0=E0c=4.083×108=1.36×10−8TE_{0} = \sqrt{\frac{2I}{\varepsilon_{0}c}} = \sqrt{\frac{2 \times \frac{2.5}{36\pi}}{\frac{1}{4\pi \times 9 \times 10^{9}} \times 3 \times 10^{8}}} = 4.08Vm^{- 1}\therefore B_{0} = \frac{E_{0}}{c} = \frac{4.08}{3 \times 10^{8}} = 1.36 \times 10^{- 8}T