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Question: The electric field of an electromagnetic wave in free space is $\vec{E}=57 \cos[7.5 \times 10^6 t - ...

The electric field of an electromagnetic wave in free space is E⃗=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)\vec{E}=57 \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (4\hat{i} - 3\hat{j}) N/C. The associated magnetic field in Tesla is-

A

B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B}=\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (5\hat{k})

B

B⃗=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec{B}=\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (\hat{k})

C

B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B}=-\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (5\hat{k})

D

B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\vec{B}=-\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (\hat{k})

Answer

B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B}=-\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (5\hat{k})

Explanation

Solution

The magnetic field B⃗\vec{B} is related to the electric field E⃗\vec{E} and the direction of propagation n^\hat{n} by the following equation:

B⃗=1cn^×E⃗\vec{B} = \frac{1}{c} \hat{n} \times \vec{E}

Where:

  • cc is the speed of light in free space (3×1083 \times 10^8 m/s)
  • n^\hat{n} is the unit vector in the direction of propagation

Given E⃗=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)\vec{E}=57 \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (4\hat{i} - 3\hat{j})

The wave vector k⃗=5×10−3(3i^+4j^)\vec{k} = 5 \times 10^{-3} (3\hat{i} + 4\hat{j}).

The unit vector in the direction of propagation is n^=k⃗∣k⃗∣=5×10−3(3i^+4j^)(5×10−3⋅3)2+(5×10−3⋅4)2=3i^+4j^32+42=3i^+4j^5\hat{n} = \frac{\vec{k}}{|\vec{k}|} = \frac{5 \times 10^{-3} (3\hat{i} + 4\hat{j})}{\sqrt{(5 \times 10^{-3} \cdot 3)^2 + (5 \times 10^{-3} \cdot 4)^2}} = \frac{3\hat{i} + 4\hat{j}}{\sqrt{3^2 + 4^2}} = \frac{3\hat{i} + 4\hat{j}}{5}.

So, n^=35i^+45j^\hat{n} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}.

Now, we compute the cross product n^×(4i^−3j^)\hat{n} \times (4\hat{i} - 3\hat{j}):

n^×(4i^−3j^)=∣i^j^k^354504−30∣=k^(35(−3)−45(4))=k^(−95−165)=−5k^\hat{n} \times (4\hat{i} - 3\hat{j}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{3}{5} & \frac{4}{5} & 0 \\ 4 & -3 & 0 \end{vmatrix} = \hat{k} \left( \frac{3}{5} (-3) - \frac{4}{5} (4) \right) = \hat{k} \left( -\frac{9}{5} - \frac{16}{5} \right) = -5\hat{k}.

Therefore, B⃗=1c⋅57cos⁡[7.5×106t−5×10−3(3x+4y)]⋅(−5k^)=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec{B} = \frac{1}{c} \cdot 57 \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] \cdot (-5\hat{k}) = -\frac{57}{3 \times 10^8} \cos[7.5 \times 10^6 t - 5 \times 10^{-3}(3x+4y)] (5\hat{k}).