The magnetic field of an electromagnetic wave is given by :- \mathop B\limits^ \to = 1.6 \times 10^ - 6\cos ( 2 \times 10^7z + 6 \times 10^15t )( 2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge )Wb \over m^2 The associated electric field will be :-

\mathop E\limits^ \to = 4.8 \times 10^2\cos ( 2 \times 10^7z + 6 \times 10^15t )( -\mathop i\limits^ \wedge + \mathop 2j\limits^ \wedge )V \over m

\mathop E\limits^ \to = 4.8 \times 10^2\cos ( 2 \times 10^7z - 6 \times 10^15t )( -2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge )V \over m

\mathop E\limits^ \to = 4.8 \times 10^2\cos ( 2 \times 10^7z - 6 \times 10^15t )( 2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge )V \over m

\mathop E\limits^ \to = 4.8 \times 10^2\cos ( 2 \times 10^7z + 6 \times 10^15t )( \mathop i\limits^ \wedge - \mathop 2j\limits^ \wedge )V \over m

Electromagnetic Waves PYQ — JEE Main 2019 · 8 Apr · Shift 2 · Q51 with Solution

View written solutionFree

Correct answer: D

Given magnetic field

\vec B = 1.6\times 10^{-6}\cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(2\hat i + \hat j)\; \text{T}

We need the associated electric field $\vec E$ of the electromagnetic wave.

Direction of propagation

For a wave,

phase $= kz-\omega t$ means propagation along $+z$ ,
phase $= kz+\omega t$ means propagation along $-z$ .

Here phase is

2\times 10^7 z + 6\times 10^{15} t

So the wave propagates along

-\hat k

Use the relation between $\vec E$ , $\vec B$ , and propagation direction

For an electromagnetic wave,

\vec B = \frac{1}{c}(\hat n \times \vec E)

where $\hat n$ is the propagation direction.

Equivalently,

\vec E = c(\vec B \times \hat n)

Here,

\hat n = -\hat k

Thus,

\vec E = c\,[\vec B \times (-\hat k)]

Now the direction part of $\vec B$ is:

2\hat i + \hat j

So,

(2\hat i + \hat j)\times (-\hat k) = -\left[(2\hat i + \hat j)\times \hat k\right]

Using

\hat i\times \hat k = -\hat j, \qquad \hat j\times \hat k = \hat i

we get

(2\hat i + \hat j)\times \hat k = 2(-\hat j)+\hat i = \hat i-2\hat j

Therefore,

(2\hat i + \hat j)\times (-\hat k)=-(\hat i-2\hat j)=-\hat i+2\hat j

So the electric field direction is

-\hat i+2\hat j

This is the same as

-(\hat i-2\hat j)

But let us also check via $\vec E\times \vec B$ must point along $-\hat k$ .

If we take direction $(\hat i-2\hat j)$ :

(\hat i-2\hat j)\times (2\hat i+\hat j) = \hat i\times \hat j -4\hat j\times \hat i = \hat k +4\hat k =5\hat k

This gives $+\hat k$ , not correct.

If we take direction $(-\hat i+2\hat j)$ :

(-\hat i+2\hat j)\times (2\hat i+\hat j) = -\hat i\times \hat j +4\hat j\times \hat i = -\hat k-4\hat k=-5\hat k

This matches propagation along $-\hat k$ .

So the correct direction is

-\hat i+2\hat j

Magnitude of electric field

For electromagnetic waves,

E_0 = cB_0

Given

B_0 = 1.6\times 10^{-6}\times \sqrt{2^2+1^2}

However, the options use the coefficient outside the vector bracket, so we compare similarly:

E_0 = c\times 1.6\times 10^{-6}

Taking

c=3\times 10^8\; \text{m/s}

E_0 = 3\times 10^8 \times 1.6\times 10^{-6} = 4.8\times 10^2\; \text{V/m}

Phase of electric field

In an electromagnetic wave, $\vec E$ and $\vec B$ are in phase. So $\vec E$ must have the same phase as $\vec B$ :

\cos\left(2\times 10^7 z + 6\times 10^{15} t\right)

Construct the electric field

Thus,

\vec E = 4.8\times 10^2 \cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(-\hat i+2\hat j)\; \text{V/m}

Among the given options, option D is written as

4.8\times 10^2 \cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(-\hat i+2\hat j)\; \text{V/m}

since

-(\hat i-2\hat j) = -\hat i+2\hat j

So the correct option is D.

Check all options briefly

A: wrong phase ( $z-\omega t$ ), wrong direction.
B: wrong phase and wrong direction.
C: correct phase, but direction $(\hat i-2\hat j)$ gives propagation along $+z$ , not $-z$ .
D: correct phase and correct direction.

Comparison with stored answer

Stored correct answer: C

My derived answer: D

So I do not agree with the stored answer. The stored answer likely missed the sign needed so that

\vec E\times \vec B

points in the propagation direction $-\hat k$ .

More from Electromagnetic Waves

View written solutionFree

Correct answer: D

Given magnetic field

\vec B = 1.6\times 10^{-6}\cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(2\hat i + \hat j)\; \text{T}

We need the associated electric field $\vec E$ of the electromagnetic wave.

Direction of propagation

For a wave,

phase $= kz-\omega t$ means propagation along $+z$ ,
phase $= kz+\omega t$ means propagation along $-z$ .

Here phase is

2\times 10^7 z + 6\times 10^{15} t

So the wave propagates along

-\hat k

Use the relation between $\vec E$ , $\vec B$ , and propagation direction

For an electromagnetic wave,

\vec B = \frac{1}{c}(\hat n \times \vec E)

where $\hat n$ is the propagation direction.

Equivalently,

\vec E = c(\vec B \times \hat n)

Here,

\hat n = -\hat k

Thus,

\vec E = c\,[\vec B \times (-\hat k)]

Now the direction part of $\vec B$ is:

2\hat i + \hat j

So,

(2\hat i + \hat j)\times (-\hat k) = -\left[(2\hat i + \hat j)\times \hat k\right]

Using

\hat i\times \hat k = -\hat j, \qquad \hat j\times \hat k = \hat i

we get

(2\hat i + \hat j)\times \hat k = 2(-\hat j)+\hat i = \hat i-2\hat j

Therefore,

(2\hat i + \hat j)\times (-\hat k)=-(\hat i-2\hat j)=-\hat i+2\hat j

So the electric field direction is

-\hat i+2\hat j

This is the same as

-(\hat i-2\hat j)

But let us also check via $\vec E\times \vec B$ must point along $-\hat k$ .

If we take direction $(\hat i-2\hat j)$ :

(\hat i-2\hat j)\times (2\hat i+\hat j) = \hat i\times \hat j -4\hat j\times \hat i = \hat k +4\hat k =5\hat k

This gives $+\hat k$ , not correct.

If we take direction $(-\hat i+2\hat j)$ :

(-\hat i+2\hat j)\times (2\hat i+\hat j) = -\hat i\times \hat j +4\hat j\times \hat i = -\hat k-4\hat k=-5\hat k

This matches propagation along $-\hat k$ .

So the correct direction is

-\hat i+2\hat j

Magnitude of electric field

For electromagnetic waves,

E_0 = cB_0

Given

B_0 = 1.6\times 10^{-6}\times \sqrt{2^2+1^2}

However, the options use the coefficient outside the vector bracket, so we compare similarly:

E_0 = c\times 1.6\times 10^{-6}

Taking

c=3\times 10^8\; \text{m/s}

E_0 = 3\times 10^8 \times 1.6\times 10^{-6} = 4.8\times 10^2\; \text{V/m}

Phase of electric field

In an electromagnetic wave, $\vec E$ and $\vec B$ are in phase. So $\vec E$ must have the same phase as $\vec B$ :

\cos\left(2\times 10^7 z + 6\times 10^{15} t\right)

Construct the electric field

Thus,

\vec E = 4.8\times 10^2 \cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(-\hat i+2\hat j)\; \text{V/m}

Among the given options, option D is written as

4.8\times 10^2 \cos\left(2\times 10^7 z + 6\times 10^{15} t\right)(-\hat i+2\hat j)\; \text{V/m}

since

-(\hat i-2\hat j) = -\hat i+2\hat j

So the correct option is D.

Check all options briefly

A: wrong phase ( $z-\omega t$ ), wrong direction.
B: wrong phase and wrong direction.
C: correct phase, but direction $(\hat i-2\hat j)$ gives propagation along $+z$ , not $-z$ .
D: correct phase and correct direction.

Comparison with stored answer

Stored correct answer: C

My derived answer: D

So I do not agree with the stored answer. The stored answer likely missed the sign needed so that

\vec E\times \vec B

points in the propagation direction $-\hat k$ .

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