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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves

एक समतल विधुत-चुम्बकीय तरंग का विधुत क्षेत्र निम्न है, \(\overrightarrow{ E }= E _{0} \hat{ i } \cos ( kz ) \cos (\omega t )\) तब संगत चुम्बकीय क्षेत्र \(\overrightarrow{ B }\) होगा :

  1. A \(\vec B = \frac{{{E_0}}}{C}\hat j\,\sin \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)\)
  2. B \(\vec B = \frac{{{E_0}}}{C}\hat k\,\sin \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)\)
  3. C \(\vec B = \frac{{{E_0}}}{C}\hat j\,\cos \,\left( {kz} \right)\,\sin \,\left( {\omega t} \right)\)
  4. D \(\vec B = \frac{{{E_0}}}{C}\hat j\,\sin \,\left( {kz} \right)\,\sin \,\left( {\omega t} \right)\)
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Answer & Solution

Correct Answer

(D) \(\vec B = \frac{{{E_0}}}{C}\hat j\,\sin \,\left( {kz} \right)\,\sin \,\left( {\omega t} \right)\)

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Detailed explanation

\(\because \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}} \| \overrightarrow{\mathrm{v}}\) Given that wave is propagating along positive \(z\) -axis and \(\overrightarrow{\mathrm{E}}\) along positive \(x\) -axis. Hence \(\overrightarrow{\mathrm{B}}\) along…
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