Question:
An electromagnetic wave travelling in the \(x\)-direction has frequency of \(2\times10^{14}\ \mathrm{Hz}\) and electric field amplitude of \(27\ \mathrm{V\,m^{-1}}\) oscillates in \(Y\)-direction. From the options given below, which one describes the magnetic field for this wave?
An electromagnetic wave travelling in the \(x\)-direction has frequency of \(2\times10^{14}\ \mathrm{Hz}\) and electric field amplitude of \(27\ \mathrm{V\,m^{-1}}\) oscillates in \(Y\)-direction. From the options given below, which one describes the magnetic field for this wave?
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B0 = E0/c, wavelength = c/f, and B must be perpendicular to both E and the propagation direction (E x B along +x).
Updated On: Nov 11, 2026
- \(\vec{B}(x,t)=(9\times10^{-8}\ \mathrm{T})\hat{j}\,\sin[1.5\times10^{-6}x-2\times10^{14}t]\)
- \(\vec{B}(x,t)=(9\times10^{-8}\ \mathrm{T})\hat{i}\,\sin[2\pi(1.5\times10^{-8}x-2\times10^{14}t)]\)
- \(\vec{B}(x,t)=(3\times10^{-8}\ \mathrm{T})\hat{j}\,\sin 2\pi\left[\left(\frac{x}{1.5\times10^{-8}}\right)-2\times10^{14}t\right]\)
- \(\vec{B}(x,t)=(9\times10^{-8}\ \mathrm{T})\hat{k}\,\sin 2\pi\left[\left(\frac{x}{1.5\times10^{-6}}\right)-2\times10^{14}t\right]\)
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The Correct Option is D
Solution and Explanation
\(B_0=\frac{E_0}{c}=\frac{27}{3\times10^{8}}=9\times10^{-8}\ \mathrm{T}\).
\(\lambda=\frac{c}{f}=\frac{3\times10^{8}}{2\times10^{14}}=1.5\times10^{-6}\ \mathrm{m}\), so the phase is \(2\pi\left(\frac{x}{\lambda}-ft\right)=2\pi\left[\frac{x}{1.5\times10^{-6}}-2\times10^{14}t\right]\).
Wave along +x with \(\vec E\) along \(\hat j\): \(\hat j\times\hat k=\hat i\), so \(\vec B\) is along \(\hat k\).
Only option (4) has amplitude \(9\times10^{-8}\) T, direction \(\hat k\) and \(\lambda=1.5\times10^{-6}\) m.
Answer: (4) \((9\times10^{-8}\,\mathrm{T})\hat k\sin2\pi\left[\frac{x}{1.5\times10^{-6}}-2\times10^{14}t\right]\)
\(\lambda=\frac{c}{f}=\frac{3\times10^{8}}{2\times10^{14}}=1.5\times10^{-6}\ \mathrm{m}\), so the phase is \(2\pi\left(\frac{x}{\lambda}-ft\right)=2\pi\left[\frac{x}{1.5\times10^{-6}}-2\times10^{14}t\right]\).
Wave along +x with \(\vec E\) along \(\hat j\): \(\hat j\times\hat k=\hat i\), so \(\vec B\) is along \(\hat k\).
Only option (4) has amplitude \(9\times10^{-8}\) T, direction \(\hat k\) and \(\lambda=1.5\times10^{-6}\) m.
Answer: (4) \((9\times10^{-8}\,\mathrm{T})\hat k\sin2\pi\left[\frac{x}{1.5\times10^{-6}}-2\times10^{14}t\right]\)
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