JEE Main · 2022 · 24 Jun 2022 · Shift 2 · EM Waves

WhatsApp Email

An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with

3.5%

efficiency.

(A) $1.19 \times 1 0^{- 8} T$
(B) $1.71 \times 1 0^{- 8} T$ ✓ Correct
(C) $0.84 \times 1 0^{- 8} T$
(D) $3.36 \times 1 0^{- 8} T$

Answer: (B)

Solution

The intensity of electromagnetic radiation at distance

r

from a point source is:

I = \frac{η P}{4 π r ^{2}} = \frac{c B _{0}^{2}}{2 μ _{0}}

Solving for peak magnetic field:

B_{0} = \frac{μ _{0} η P}{2 π r ^{2} c}

Substituting values:

η = 0.035

P = 200

r = 4

c = 3 \times 1 0^{8}

m/s,

μ_{0} = 4 π \times 1 0^{- 7}

T·m/A:

B_{0} = \frac{1}{4} \frac{1 0 ^{- 7} \times 4 \times 3.5}{3 \times 1 0 ^{8}} = 1.71 \times 1 0^{- 8}

T

Answer: (B)

1.71 \times 1 0^{- 8} T

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on EM Waves →

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · EM Waves

Solution

More JEE Main questions on EM Waves