Question

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light intensity is Io, the average energy density of the waves is u0, and the rms electric and magnetic field values are Eo and Bo, respectively.
Required:
a. What is the light intensity?
b. What is the average energy density of the waves?
c. What is the rms magnetic field value?

Answer 1

Answer:

The appropriate solution is:

(a) $\frac{1}{4}(I_o)$

(b) $\frac{1}{4} (u_o)$

(c) $\frac{1}{2}B_o$

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒  $\frac{I}{I_o} =\frac{r_o_2}{r_2}$

On applying cross-multiplication, we get

⇒  $I=I_o\times \frac{1_2}{2^2}$

⇒     $=I_o\times \frac{1}{4}$

⇒     $=\frac{1}{4} (I_o)$

(b)

As we know,

⇒ $\frac{u}{u_o} =\frac{I}{I_o}$

By putting the values, we get

⇒ $u=\frac{1}{4}(u_o)$

(c)

⇒ $\frac{B^2}{B_o^2} =\frac{u}{u_o}$

         $=\frac{I}{I_o}$

⇒ $B=B_o\times \sqrt{\frac{1}{4} }$

⇒     $=\frac{1}{2}(B_o)$