Question

Calculate the electric and magnetic fields produced by radiation coming from a 100 w bulb at a distance of 3 m . assume the efficiency of bulb is 2.5 percent and it is a source point

Answer 1

Answer:

$E_{0} =4.07Volt\ m^{-1}$

$E_{0} =1.4\times 10^{-8}Tesla$

Explanation:

As per the question,

Given:

Power = 100 W

Efficiency = 2.5 % = 0.025

Radius = 3 m

Now

The surface area of surrounding of sphere is given by

A = 4 πr²

⇒ A = 4 × 3.14 × 3² = 113 m²

And the intensity at this radius is given by

$I=\frac{Power\times efficiency}{Area}$

$I=\frac{100\times 0.025}{113}$

I = 0.002 W/m²

As it is given that total intensity is half provided by the electric field and half by the magnetic field.

∴ $\frac{1}{2}I=\frac{1}{2}(\epsilon_{0}E_{rms}^{2}c)$

c = speed of light = 3 × 10⁸ ms⁻¹

That is,

$E_{rms}=\sqrt{\frac{0.002}{8.85\times 10^{-12}\times 3\times 10^{8}}}$

$E_{rms}=2.9 Volt\ m^{-1}$

As the electric field in a light beam is of the form of sinusoidal, then the peak electric field is given by

$E_{0} = \sqrt{2} E_{rms}$

$E_{0} = \sqrt{2}\times 2.9$

$E_{0} =4.07Volt\ m^{-1}$

Now,

we know that,

$B_{rms}=\frac{E_{rms}}{c}$

∴ $B_{rms}=\frac{2.9}{3\times 10^{8}}$

   $B_{rms}=9.6\times 10^{-9}Tesla$

Also,

As the magnetic field in a light beam is of the form of sinusoidal, then the peak magnetic field is given by

$B_{0} = \sqrt{2} B_{rms}$

$B_{0} = \sqrt{2}\times 9.6\times 10^{-9}$

$E_{0} =1.4\times 10^{-8}Tesla$

Answer 2

Answer:

E_{0} =4.07Volt\ m^{-1}E

0

=4.07Volt m

−1

E_{0} =1.4\times 10^{-8}TeslaE

0

=1.4×10

−8

Tesla

Explanation:

As per the question,

Given:

Power = 100 W

Efficiency = 2.5 % = 0.025

Radius = 3 m

Now

The surface area of surrounding of sphere is given by

A = 4 πr²

⇒ A = 4 × 3.14 × 3² = 113 m²

And the intensity at this radius is given by

I=\frac{Power\times efficiency}{Area}I=

Area

Power×efficiency

I=\frac{100\times 0.025}{113}I=

113

100×0.025

I = 0.002 W/m²

As it is given that total intensity is half provided by the electric field and half by the magnetic field.

∴ \frac{1}{2}I=\frac{1}{2}(\epsilon_{0}E_{rms}^{2}c)

2

1

I=

2

1

(ϵ

0

E

rms

2

c)

c = speed of light = 3 × 10⁸ ms⁻¹

That is,

E_{rms}=\sqrt{\frac{0.002}{8.85\times 10^{-12}\times 3\times 10^{8}}}E

rms

=

8.85×10

−12

×3×10

8

0.002

E_{rms}=2.9 Volt\ m^{-1}E

rms

=2.9Volt m

−1

As the electric field in a light beam is of the form of sinusoidal, then the peak electric field is given by

E_{0} = \sqrt{2} E_{rms}E

0

=

2

E

rms

E_{0} = \sqrt{2}\times 2.9E

0

=

2

×2.9

E_{0} =4.07Volt\ m^{-1}E

0

=4.07Volt m

−1

Now,

we know that,

B_{rms}=\frac{E_{rms}}{c}B

rms

=

c

E

rms

∴ B_{rms}=\frac{2.9}{3\times 10^{8}}B

rms

=

3×10

8

2.9

B_{rms}=9.6\times 10^{-9}TeslaB

rms

=9.6×10

−9

Tesla

Also,

As the magnetic field in a light beam is of the form of sinusoidal, then the peak magnetic field is given by

B_{0} = \sqrt{2} B_{rms}B

0

=

2

B

rms

B_{0} = \sqrt{2}\times 9.6\times 10^{-9}B 0 =2×9.6×10 −9

E_{0} =1.4\times 10^{-8}TeslaE0

=1.4×10−8

Tesla