A plane electromagnetic wave is moving in free space with velocity c = 3 × 10 8 m/s and its electric field is given as is the unit vector along y-axis. The magnetic field vector B of the wave is: [2026]

Question

A plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^{8} m/s$ and its electric field is given as $\vec{E} = 54 \sin (k z - ω t) \hat{j} V/m,$ where $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is: [2026]

Smart Achievers · Accepted Answer

(1)

$\hat{B} = \hat{C} \times \hat{E} = \hat{k} \times \hat{j} = - \hat{i}$

$∴$ $\vec{B} = \frac{54}{3 \times 10^{8}} \sin (k z - ω t) (- \hat{i})$

$= - 1.8 \times 10^{- 7} \sin (k z - ω t) \hat{i}$

Solution

Q.
A plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^{8} m/s$ and its electric field is given as $\vec{E} = 54 \sin (k z - ω t) \hat{j} V/m,$ where $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is: [2026]

1 $- 1.8 \times 10^{- 7} \sin (k z - ω t) \hat{i} T$

2 $1.4 \times 10^{- 7} \sin (k z - ω t) \hat{i} T$

3 $1.4 \times 10^{- 7} \sin (k z - ω t) \hat{k} T$

4 $+ 1.8 \times 10^{- 7} \sin (k z - ω t) \hat{i} T$

Ans.
(1)

$\hat{B} = \hat{C} \times \hat{E} = \hat{k} \times \hat{j} = - \hat{i}$

$∴$ $\vec{B} = \frac{54}{3 \times 10^{8}} \sin (k z - ω t) (- \hat{i})$

$= - 1.8 \times 10^{- 7} \sin (k z - ω t) \hat{i}$

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Solution

Q. A plane electromagnetic wave is moving in free space with velocity c=3×108 m/s and its electric field is given as E→=54sin(kz-ωt)j^ V/m, where j^ is the unit vector along y-axis. The magnetic field vector B→ of the wave is: [2026]

1 -1.8×10-7sin(kz-ωt)i^T

2 1.4×10-7sin(kz-ωt)i^T

3 1.4×10-7sin(kz-ωt)k^T

4 +1.8×10-7sin(kz-ωt)i^T

Ans. (1) B^=C^×E^=k^×j^=-i^ ∴ B→=543×108sin(kz-ωt)(-i^) =-1.8×10-7sin(kz-ωt)i^