- JEE Mains 2022
- Medium
- MCQ
If electric field intensity of a uniform plane electro magnetic wave is given as \(E =-301.6 \sin ( kz -\omega t ) \hat{a}_{ x }+452.4 \sin ( kz -\omega t )\) \(\hat{a}_{y} \frac{V}{m}\) Then, magnetic intensity \(H\) of this wave in \(Am ^{-1}\) will be [Given: Speed of light in vacuum \(c =3 \times 10^{8} ms ^{-1}\), permeability of vacuum \(\mu_{0}=4 \pi \times 10^{-7} NA ^{-2}\) ]
Answer & Solution
(C) \(-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ B }=\frac{301.6}{ C } \sin ( kz -\omega t )\left(-\hat{a}_{ y }\right)+\frac{452.4}{ C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)\)
\(\overrightarrow{ H }=\frac{\overrightarrow{ B }}{\mu_{0}}=\frac{301.6}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ y }\right)+\frac{452.4}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)\)
\(\overrightarrow{ H }=-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }\)
For direction
\(\overrightarrow{ E } \times \overrightarrow{ B }\) is direction of \(\overrightarrow{ C }\)
For first part \(\hat{ E }=-\hat{ i }, \hat{ B }= ?\)
\(\hat{ E } \times \hat{ B }=\hat{ k } \Rightarrow \hat{ B }=-\hat{ j }\)
Similarly for second
\(\hat{ E }=\hat{ j }, \hat{ B }=?\)
\(\hat{ E } \times \hat{ B }=\hat{ k } \Rightarrow \hat{ B }=-\hat{ i }\)
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