JEE Main 2022 · 30 Jun · Shift 1 · Physics Q44

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PhysicsSingle correct+4 / -1Question 44 of 90

An expression for oscillating electric field in a plane electromagnetic wave is given as $E_z=300\sin\left(5\pi\times10^3x-3\pi\times10^{11}t\right)$ Vm $^{-1}$ . Then, the value of magnetic field amplitude will be :

(Given : speed of light in Vacuum $c=3\times10^8$ ms $^{-1}$ )

1\times10^{-6}

5\times10^{-6}

18\times10^9

21\times10^9

Answer and worked solution

Official answer: B

Wave speed from the equation: $\frac\omega k=\frac{3\pi\times10^{11}}{5\pi\times10^3}=6\times10^7$ m s $^{-1}$ , so $B_0=\frac{300}{6\times10^7}=5\times10^{-6}$ T, the keyed value. (Using $c$ instead would give $10^{-6}$ T.)

(B)

Topics:Properties of EM waves

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JEE Main 2022 · 30 Jun · Shift 1, Physics question 44