The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is x \over 10\sqrt \mu _0c \over \pi V \over m . The efficiency of the bulb is 10% and it is a point source. The value of x is \underline\hspace2cm .

Electromagnetic Waves PYQ — JEE Main 2021 · 25 Feb · Shift 2 · Q71 with Solution

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Correct answer: X = \DFRAC{2}{\SQRT{10}}

Useful power emitted as radiation

The bulb is rated at $8\,\text{W}$ , but its efficiency is $10\%$ . So the electromagnetic power actually radiated is

P = 0.1 \times 8 = 0.8\,\text{W}.

Intensity at distance $r=10\,\text{m}$

Since the bulb is a point source, radiation spreads uniformly over a sphere of radius $r$ . Thus,

I = \frac{P}{4\pi r^2}.

Substituting values,

I = \frac{0.8}{4\pi (10)^2} = \frac{0.8}{400\pi} = \frac{1}{500\pi}\,\text{W/m}^2.

Relation between intensity and peak electric field

For an electromagnetic wave,

I = \frac{1}{2}\,\varepsilon_0 c E_0^2.

Using $\varepsilon_0 = \dfrac{1}{\mu_0 c^2}$ , this becomes

I = \frac{E_0^2}{2\mu_0 c}.

Hence,

E_0 = \sqrt{2\mu_0 c I}.

Now substitute $I = \dfrac{1}{500\pi}$ :

E_0 = \sqrt{2\mu_0 c \cdot \frac{1}{500\pi}} = \sqrt{\frac{\mu_0 c}{250\pi}}.

Rewrite:

E_0 = \frac{1}{\sqrt{250}}\sqrt{\frac{\mu_0 c}{\pi}} = \frac{1}{5\sqrt{10}}\sqrt{\frac{\mu_0 c}{\pi}}.

Compare with the given form

Given,

E_0 = \frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}}\,\text{V/m}.

So,

\frac{x}{10} = \frac{1}{5\sqrt{10}}.

Therefore,

x = \frac{2}{\sqrt{10}}.

This is not an integer, so likely the intended field is the rms electric field instead of the peak field, or the printed expression is meant to simplify numerically.

Let us check the common JEE convention using

I = \frac{E^2}{\mu_0 c}

for field amplitude as written in some problems. Then,

E = \sqrt{\mu_0 c I} = \sqrt{\mu_0 c \cdot \frac{1}{500\pi}} = \frac{1}{10\sqrt{5}}\sqrt{\frac{\mu_0 c}{\pi}},

which still does not give an integer $x$ .

Now, if the denominator in the intensity relation is taken as $4\pi r^2$ and the expression is expected in the form

\frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}},

then matching the stored answer $x=2$ corresponds to

E_0 = \frac{2}{10}\sqrt{\frac{\mu_0 c}{\pi}} = \frac{1}{5}\sqrt{\frac{\mu_0 c}{\pi}},

which would arise if one effectively uses

I = \frac{1}{50\pi}.

That does not match the stated data.

Conclusion

Using the standard electromagnetic-wave relation for peak electric field, the result is

x = \frac{2}{\sqrt{10}},

which is not an integer. Therefore the stored answer appears inconsistent with the given statement.

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View written solutionFree

Correct answer: X = \DFRAC{2}{\SQRT{10}}

Useful power emitted as radiation

The bulb is rated at $8\,\text{W}$ , but its efficiency is $10\%$ . So the electromagnetic power actually radiated is

P = 0.1 \times 8 = 0.8\,\text{W}.

Intensity at distance $r=10\,\text{m}$

Since the bulb is a point source, radiation spreads uniformly over a sphere of radius $r$ . Thus,

I = \frac{P}{4\pi r^2}.

Substituting values,

I = \frac{0.8}{4\pi (10)^2} = \frac{0.8}{400\pi} = \frac{1}{500\pi}\,\text{W/m}^2.

Relation between intensity and peak electric field

For an electromagnetic wave,

I = \frac{1}{2}\,\varepsilon_0 c E_0^2.

Using $\varepsilon_0 = \dfrac{1}{\mu_0 c^2}$ , this becomes

I = \frac{E_0^2}{2\mu_0 c}.

Hence,

E_0 = \sqrt{2\mu_0 c I}.

Now substitute $I = \dfrac{1}{500\pi}$ :

E_0 = \sqrt{2\mu_0 c \cdot \frac{1}{500\pi}} = \sqrt{\frac{\mu_0 c}{250\pi}}.

Rewrite:

E_0 = \frac{1}{\sqrt{250}}\sqrt{\frac{\mu_0 c}{\pi}} = \frac{1}{5\sqrt{10}}\sqrt{\frac{\mu_0 c}{\pi}}.

Compare with the given form

Given,

E_0 = \frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}}\,\text{V/m}.

So,

\frac{x}{10} = \frac{1}{5\sqrt{10}}.

Therefore,

x = \frac{2}{\sqrt{10}}.

This is not an integer, so likely the intended field is the rms electric field instead of the peak field, or the printed expression is meant to simplify numerically.

Let us check the common JEE convention using

I = \frac{E^2}{\mu_0 c}

for field amplitude as written in some problems. Then,

E = \sqrt{\mu_0 c I} = \sqrt{\mu_0 c \cdot \frac{1}{500\pi}} = \frac{1}{10\sqrt{5}}\sqrt{\frac{\mu_0 c}{\pi}},

which still does not give an integer $x$ .

Now, if the denominator in the intensity relation is taken as $4\pi r^2$ and the expression is expected in the form

\frac{x}{10}\sqrt{\frac{\mu_0 c}{\pi}},

then matching the stored answer $x=2$ corresponds to

E_0 = \frac{2}{10}\sqrt{\frac{\mu_0 c}{\pi}} = \frac{1}{5}\sqrt{\frac{\mu_0 c}{\pi}},

which would arise if one effectively uses

I = \frac{1}{50\pi}.

That does not match the stated data.

Conclusion

Using the standard electromagnetic-wave relation for peak electric field, the result is

x = \frac{2}{\sqrt{10}},

which is not an integer. Therefore the stored answer appears inconsistent with the given statement.

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