JEE Main 2015 · 10 Apr · Shift 1 · Physics Q23

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PhysicsSingle correct+4 / -1Question 23 of 90

An electromagnetic wave travelling in the $x$ -direction has frequency of $2\times10^{14}$ Hz and electric field amplitude of 27 Vm $^{-1}$ . From the options given below, which one describes the magnetic field for this wave ?

\vec B(x,t)=(3\times10^{-8}\,\text{T})\hat j\sin\left[2\pi(1.5\times10^{-8}x-2\times10^{14}t)\right]

\vec B(x,t)=(9\times10^{-8}\,\text{T})\hat k\sin\left[2\pi(1.5\times10^{-6}x-2\times10^{14}t)\right]

\vec B(x,t)=(9\times10^{-8}\,\text{T})\hat i\sin\left[2\pi(1.5\times10^{-8}x-2\times10^{14}t)\right]

\vec B(x,t)=(9\times10^{-8}\,\text{T})\hat j\sin\left[1.5\times10^{-6}x-2\times10^{14}t\right]

Answer and worked solution

Official answer: Bonus: full marks to everyone

$B_0=\frac{E_0}c=9\times10^{-8}$ T, and $\frac1\lambda=\frac fc=\frac{2\times10^{14}}{3\times10^8}=6.7\times10^5$ m⁻¹. For a wave along $x$ , $\vec B$ must lie along $\hat j$ or $\hat k$ .

None of the options has all three right, and the question was given marks to all.

Topics:Properties of EM waves

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 23