JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
Light wave traveling in air along \(x\)-direction is given by \(E _{ y }=540 \sin \pi \times 10^{4}( x - ct ) Vm ^{-1}\). Then, the peak value of magnetic field of wave will be \(\dots \times 10^{-7}\,T\) (Given \(c =3 \times 10^{8}\,ms ^{-1}\) )
- A \(18\)
- B \(54\)
- C \(5.4\)
- D \(1.8\)
Answer & Solution
Correct Answer
(A) \(18\)
Step-by-step Solution
Detailed explanation
\(E _{ y }=540 \sin \pi \times 10^{4}( x - ct ) Vm ^{-1}\) \(E _{0}=540\,Vm ^{-1}\) \(B _{0}=\frac{ E _{0}}{ C }=\frac{540}{3 \times 10^{8}}=18 \times 10^{-7}\,T\)
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