JEE Main · 2025 · 28 Jan 2025 · Shift 1 · EM Waves

Due to presence of an em-wave whose electric component is given by

E = 100 sin (ω t - kx) NC^{- 1}

, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

(A) $25 sin (ω t - kx) NC^{- 1}$
(B) $200 sin (ω t - kx) NC^{- 1}$ ✓ Correct
(C) $400 sin (ω t - k x) NC^{- 1}$
(D) $50 sin (ω t - kx) NC^{- 1}$

Answer: (B)

Solution

The energy density of an electromagnetic wave is proportional to

E^{2}

, so the total energy contained in a cylindrical volume of radius r and length L is

U \propto E^{2} π r^{2} L

Both cylinders have the same length and hold equal total energy:

E_{1}^{2} r_{1}^{2} = E_{2}^{2} r_{2}^{2} ⟹ E_{1} r_{1} = E_{2} r_{2}

With

r_{2} = r_{1} /2

and

E_{1} = 100 NC^{- 1}

E_{2} = E_{1} \cdot \frac{r _{1}}{r _{2}} = 100 \times 2 = 200 NC^{- 1}

So the electric field of the second wave must be

200 sin (ω t - k x) NC^{- 1}

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on EM Waves →

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · EM Waves

Solution

More JEE Main questions on EM Waves