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

વિદ્યુત ચુંબકીય તરંગમાં વિદ્યુતક્ષેત્ર \(\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right) N \mathrm{NC}^{-1}\) થી આપવામાં આવે છે. આ તરંગમાં ચુંબકીય ક્ષેત્ર _______ થશે.

  1. A \(\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\)
  2. B \(\vec{B}=\hat{j} 40 \cos \omega\left(t-\frac{z}{c}\right)\)
  3. C \(\overrightarrow{\mathrm{B}}=\hat{\mathrm{k}} \frac{40}{\mathrm{c}} \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\)
  4. D \(\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\vec{B}=\hat{j} \frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right)\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\) \(\overrightarrow{\mathrm{E}}\) is along \(+\mathrm{x}\) direction \(\overrightarrow{\mathrm{v}}\) is along \(+\mathrm{z}\) direction So direction of \(\vec{B}\)…
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