JEE Main 2022 · 24 Jun · Shift 2 · Physics Q47

← JEE Main 2022 · 24 Jun · Shift 2

PhysicsSingle correct+4 / -1Question 47 of 90

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.

1.19\times10^{-8}

1.71\times10^{-8}

0.84\times10^{-8}

3.36\times10^{-8}

Answer and worked solution

Official answer: B

Radiated power $0.035\times200=7$ W, so at 4 m $I=\dfrac7{4\pi(16)}\approx0.0348$ W m $^{-2}$ .

Peak field. $I=\dfrac{cB_0^2}{2\mu_0}$ , so $B_0=\sqrt{\dfrac{2\mu_0I}c}=\sqrt{\dfrac{2(4\pi\times10^{-7})(0.0348)}{3\times10^8}}\approx1.71\times10^{-8}$ T.

(B)

Topics:Energy and momentum of EM waves

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 47