The magnetic field of an electromagnetic wave is given by:${\rm{\vec B}} = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}$ The associated electric field will be:

Question

The magnetic field of an electromagnetic wave is given by:${\rm{\vec B}} = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}$ The associated electric field will be:

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TuteeHUB · Answer

${\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}$

TuteeHUB · Answer

${\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( { - 2\hat j + \hat i} \right)\frac{{\rm{V}}}{{\rm{m}}}$

TuteeHUB · Answer

${\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( { - \hat i + 2\hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}$

TuteeHUB · Answer

${\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {\hat i - 2\hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}$

1.	The magnetic field of an electromagnetic wave is given by:\({\rm{\vec B}} = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}\) The associated electric field will be:
A.	\({\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}\)
B.	\({\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} - 6 \times {{10}^{15}}{\rm{t}}} \right)\left( { - 2\hat j + \hat i} \right)\frac{{\rm{V}}}{{\rm{m}}}\)
C.	\({\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( { - \hat i + 2\hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}\)
D.	\({\rm{\vec E}} = 4.8 \times {10^2}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {\hat i - 2\hat j} \right)\frac{{\rm{V}}}{{\rm{m}}}\)
Answer» E.

The magnetic field of an electromagnetic wave is given by:\({\rm{\vec B}} = 1.6 \times {10^{ - 6}}{\rm{cos}}\left( {2 \times {{10}^7}{\rm{z}} + 6 \times {{10}^{15}}{\rm{t}}} \right)\left( {2\hat i + \hat j} \right)\frac{{{\rm{Wb}}}}{{{{\rm{m}}^2}}}\) The associated electric field will be:

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