સમતલ વિદ્યુતચુંબકીય તરંગ માટે વિદ્યુ

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8.Electromagnetic waves

medium

સમતલ વિદ્યુતચુંબકીય તરંગ માટે વિદ્યુતક્ષેત્ર $E =-301.6 \sin ( k z-\omega t ) \hat{ a }_{x}+452.4 \sin ( k z-\omega t ) \hat{ a }_{y}\, \frac{ V }{ m }$ વડે આપવામાં આવે છે. આ તરંગ માટે ચુંબકીય ક્ષેત્રની તીવ્રતા ..........વડે આપી શકાય.

[આપેલ : પ્રકાશની ઝડપ $c =3 \times 10^{8} \,ms ^{-1}$, શુન્યાવકાશની પરમીએબિલીટી $\mu_{0}=4 \pi \times 10^{-7} \,NA ^{-2}$]

$+0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }+0.8 \sin ( kz -\omega t ) \hat{ a }_{ x }$

$+1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }+1.5 \times 10^{-6}( kz -\omega t ) \hat{ a }_{ x }$

$-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }$

$-1.0 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ y }-1.5 \times 10^{-6} \sin ( kz -\omega t ) \hat{ a }_{ x }$

(JEE MAIN-2022)

Solution

$\overrightarrow{ E }=301.6 \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)+452.4 \sin ( kz -\omega t ) \hat{ a }_{ y }$

$\overrightarrow{ B }=\frac{301.6}{ C } \sin ( kz -\omega t )\left(-\hat{a}_{ y }\right)+\frac{452.4}{ C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)$

$\overrightarrow{ H }=\frac{\overrightarrow{ B }}{\mu_{0}}=\frac{301.6}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ y }\right)+\frac{452.4}{\mu C } \sin ( kz -\omega t )\left(-\hat{ a }_{ x }\right)$

$\overrightarrow{ H }=-0.8 \sin ( kz -\omega t ) \hat{ a }_{ y }-1.2 \sin ( kz -\omega t ) \hat{ a }_{ x }$

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

$\hat{ E }=\hat{ j }, \hat{ B }=?$

$\hat{ E } \times \hat{ B }=\hat{ k } \Rightarrow \hat{ B }=-\hat{ i }$

Standard 12

Physics