Question

The magnetic field of an electromagnetic wave is given by:
⃗B = 1.6 ×10⁻⁶ cos (2 × 10⁷z + 6 × 10¹⁵ t)(2î + jˆ)Wb/2²
The associated electric field will be:
(A) ⃗E = 4.8 ×10² cos (2 × 10⁷z + 6 × 10¹⁵ t)(-î + 2jˆ)V/m
(B) ⃗E = 4.8 ×10² cos (2 × 10⁷z + 6 × 10¹⁵ t)(-2jˆ + 2tˆ)V/m
(C) ⃗E = 4.8 ×10² cos (2 × 10⁷z + 6 × 10¹⁵ t)(î + 2jˆ)V/m
(D) ⃗E = 4.8 ×10² cos (2 × 10⁷z + 6 × 10¹⁵ t)(2î + jˆ)V/m

Answer 1

Answer:

correct answer c

Explanation:

I it's help u

Answer 2

The associated electric field is given by$\vec E = 4.8 \times 10^{2} \cos (2 \times 10^{7}z + 6 \times 10^{15}t ) (-\vec i + 2 \vec j) \frac{V}{m}$

Option (A) is correct.

Explanation:

Given :

Magnetic field  $\vec B = 1.6 \times 10^{-6} \cos (2 \times 10^{7}z + 6 \times 10^{15}t ) (2 \vec i +\vec j)$

We know that the electromagnetic waves is transverse wave means amplitude of waves is perpendicular to the direction of propagation.

⇒ $\vec E$ ⊥ $\vec B$ and both are ⊥ $\vec n$.

Where $\vec n =$ propagation vector.

In our case, wave propagate in $(- z)$ direction means $-\vec k$ direction.

From EM-waves theory,

  $\vec E = -c ( \vec n \times \vec B)$

Where $c = 3 \times 10^{8}$

 $\vec E = 4.8 \times 10^{2} \cos (2 \times 10^{7}z + 6 \times 10^{15}t ) [ \vec k \times (2 \vec i+ \vec j)]$

Since $\vec k \times 2 \vec i = 2 \vec j$ and $\vec k \times \vec j = - \vec i$

So we can write

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