JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2015
- Medium
- MCQ
An electromagnetic wave travelling in the \(x-\) direction has frequency of \(2 \times 10^{14}\,Hz\) and electric field amplitude of \(27\,Vm^{-1}\) . From the options given below, which one describes the magnetic field for this wave ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\vec B\,(x\,,\,t) = (9 \times {10^{ - 8}}\,T)\,\hat k \;\sin \,[2\pi \,(1.5 \times {10^{ - 6}}\,x\, - \,2 \times {10^{14}}\,t)]\)
Step-by-step Solution
Detailed explanation
As we know,
\(B_{0}=\frac{E_{0}}{C}=\frac{27}{3 \times 10^{8}}=9 \times 10^{-8}\) \(tesla\)
Oscillation of \(B\) can be only along \(\hat{\jmath}\) or \(\hat{k}\) direction.
\(\omega=2 \pi f=2 \pi \times 2 \times 10^{14} \,\mathrm{Hz}\)
\(\therefore \bar{B}(x, t)=\left(9 \times 10^{-8} T\right) \hat{k} \sin [2 \pi(1)\left.1.5 \times 10^{-6} \times-2 \times 10^{4} t\right)\)
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