The electric field of an electromagnetic wave in free space is The associated magnetic field in Tesla is: [2025]

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Q.
The electric field of an electromagnetic wave in free space is

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

The associated magnetic field in Tesla is: [2025]

1 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

2 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$

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

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

Ans.
(3)

E = CB

B = $\frac{57 \times 5}{3 \times 10^{8}}$

$3 \hat{i} + 4 \hat{j}$ is direction of propagation

$\Rightarrow (4 \hat{i} - 3 \hat{j}) \times (- \hat{k}) = 4 \hat{j} + 3 \hat{i}$

Solution

Q.
The electric field of an electromagnetic wave in free space is

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

The associated magnetic field in Tesla is: [2025]

1 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

2 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$

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

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

Ans.
(3)

E = CB

B = $\frac{57 \times 5}{3 \times 10^{8}}$

$3 \hat{i} + 4 \hat{j}$ is direction of propagation

$\Rightarrow (4 \hat{i} - 3 \hat{j}) \times (- \hat{k}) = 4 \hat{j} + 3 \hat{i}$

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Solution

Q. The electric field of an electromagnetic wave in free space is E→=57 cos [7.5×106t–5×10–3(3x+4y)](4i^–3j^)N/C The associated magnetic field in Tesla is: [2025]

1 B→=573×108 cos [7.5×106t–5×10–3(3x+4y)](5k^)

2 B→=573×108 cos [7.5×106t–5×10–3(3x+4y)](k^)

3 B→=–573×108 cos [7.5×106t–5×10–3(3x+4y)](5k^)

4 B→=–573×108 cos [7.5×106t–5×10–3(3x+4y)](k^)

Ans. (3) E = CB B = 57×53×108 3i^+4j^ is direction of propagation ⇒ (4i^–3j^)×(–k^)=4j^+3i^

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Q.
The electric field of an electromagnetic wave in free space is

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

The associated magnetic field in Tesla is: [2025]

1 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (5 \hat{k})$

2 $\vec{B} = \frac{57}{3 \times 10^{8}} \cos [7.5 \times 10^{6} t - 5 \times 10^{- 3} (3 x + 4 y)] (\hat{k})$

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

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

Ans.
(3)

E = CB

B = $\frac{57 \times 5}{3 \times 10^{8}}$

$3 \hat{i} + 4 \hat{j}$ is direction of propagation

$\Rightarrow (4 \hat{i} - 3 \hat{j}) \times (- \hat{k}) = 4 \hat{j} + 3 \hat{i}$