JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The magnetic field of an electromagnetic wave is given by \(\vec B = 1.6 \times {10^{ - 6}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\hat i + \hat j} \right)\frac{{Wb}}{{{m^2}}}\) The associated electric field will be
- A \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - \hat i + 2\hat j} \right)\frac{V}{m}\)
- B \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - 2\hat j + 2\hat i} \right)\frac{V}{m}\)
- C \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\hat i + 2\hat j} \right)\frac{V}{m}\)
- D \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\hat i + \hat j} \right)\frac{V}{m}\)
Answer & Solution
Correct Answer
(A) \(\vec E = 4.8 \times {10^{ 2}}\,\cos \,\left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( { - \hat i + 2\hat j} \right)\frac{V}{m}\)
Step-by-step Solution
Detailed explanation
If we use that direction of light propagation will be along \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\). Then \((\mathrm{A})\) option is correct. Magnitude of \(E=C B\) \(E=3 \times 10^{8} \times 1.6 \times 10^{-6} \times \sqrt{5}\)…
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