The electric field of an electromagnetic wave in free space is E = 57 [7.5 10^6 t - 5 10^-3 (3x + 4y) ] ( 4 i - 3 j) N/C. The associated magnetic field in Tesla is:

Given: The electric field is E = 57 [7.5 10^6 t - 5 10^-3 (3x + 4y) ] (4 i - 3 j) N/C Find: The associated magnetic field B. For an electromagnetic wave in free space, the magnetic field is perpendicular to both the electric field and the direction of propagation. The propagation direction is obtained from the phase term 5 10^-3(3x+4y), so the wave vector direction is along 3 i+4 j. The corresponding unit vector along propagation is k wave = 3 i+4 j 3^2+4^2 = 3 i+4 j5 Now evaluate the direction using the cross product with the electric field direction: k wave (4 i-3 j) = vmatrix i & j & k 35 & 45 & 0 4 & -3 & 0 vmatrix This gives k wave (4 i-3 j) = ( 35 (-3) - 45 4) k = - ( 95+ 165) k = -5 k

The electric field of an electromagnetic wave in free space is E… — Solution | JEE Physics

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

A
$\vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k})$
B
$\vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k})$
C
$\vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k})$
D
$\vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k})$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The electric field is

\vec{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (4\hat{i} - 3\hat{j}) \, \text{N/C}

Find: The associated magnetic field $\vec{B}$ .

For an electromagnetic wave in free space, the magnetic field is perpendicular to both the electric field and the direction of propagation. The propagation direction is obtained from the phase term $5 \times 10^{-3}(3x+4y)$ , so the wave vector direction is along $3\hat{i}+4\hat{j}$ .

The corresponding unit vector along propagation is

\hat{k}_{\text{wave}} = \frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}} = \frac{3\hat{i}+4\hat{j}}{5}

Cross Product Evaluation

Now evaluate the direction using the cross product with the electric field direction:

\hat{k}_{\text{wave}} \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}

This gives

\hat{k}_{\text{wave}} \times (4\hat{i}-3\hat{j}) = \left(\frac{3}{5}\times(-3) - \frac{4}{5}\times 4\right)\hat{k} = -\left(\frac{9}{5}+\frac{16}{5}\right)\hat{k} = -5\hat{k}

Common mistakes

Using only $|\vec{B}| = \frac{|\vec{E}|}{c}$ and ignoring direction is incorrect, because the magnetic field must also be perpendicular to both $\vec{E}$ and the propagation direction. After finding the magnitude, determine the vector direction with the cross product.
Taking the propagation direction directly as $4\hat{i}-3\hat{j}$ is wrong, because that vector gives the direction of the electric field, not the wave vector. The propagation direction comes from the spatial part of the phase, namely $3\hat{i}+4\hat{j}$ .
Forgetting to normalize $3\hat{i}+4\hat{j}$ leads to a wrong factor. The unit propagation vector is $\frac{3\hat{i}+4\hat{j}}{5}$ , and this normalization produces the factor $5\hat{k}$ in the final answer.

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Answer

Step-by-step solution

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