The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression \overrightarrow E ( x,y ) = 10\widehat j\cos [ ( 6x + 8z ) ]. The magnetic field \overrightarrow B (x,z, t) is given by - (c is the velocity of light)

1 \over c( 6\widehat k - 8\widehat i )\cos [ ( 6x + 8z - 10ct ) ]

1 \over c( 6\hat k + 8\widehat i )\cos [ ( 6x + 8z - 10ct ) ]

1 \over c( 6\hat k + 8\widehat i )\cos [ ( 6x - 8z + 10ct ) ]

1 \over c( 6\hat k - 8\widehat i )\cos [ ( 6x - 8z + 10ct ) ]

Electromagnetic Waves PYQ — JEE Main 2019 · 10 Jan · Shift 2 · Q65 with Solution

View written solutionFree

Correct answer: B

Given electric field at $t=0$

$\vec E(x,z,0)=10\,\hat j\cos(6x+8z)$

For a plane electromagnetic wave, the general form is

$\vec E=\vec E_0\cos(\vec k\cdot \vec r-\omega t)$

or equivalently with a plus sign depending on propagation direction.
Identify the wave vector

From

$\cos(6x+8z)$

we get

$\vec k=6\hat i+8\hat k$

and its magnitude is

$|\vec k|=\sqrt{6^2+8^2}=10$

Hence

$\omega=c|\vec k|=10c$

So the time-dependent phase must be

$6x+8z-10ct$

for propagation along $+\vec k$ .
Direction of magnetic field

For an electromagnetic wave in free space,

$\vec B=\frac{1}{\omega}\,\vec k\times \vec E$

or equivalently,

$\vec B=\frac{1}{c}\,\hat n\times \vec E$

where $\hat n=\dfrac{\vec k}{|\vec k|}$ is the propagation direction.

Here,
$\qquad \vec k=6\hat i+8\hat k$$ Compute cross product: $$\vec k\times \vec E_0=(6\hat i+8\hat k)\times 10\hat j$$ $$=10\left[6(\hat i\times \hat j)+8(\hat k\times \hat j)\right]$$ Using $$\hat i\times \hat j=\hat k, \qquad \hat k\times \hat j=-\hat i$$ we get $$\vec k\times \vec E_0=10(6\hat k-8\hat i)$$ Therefore, $$\vec B=\frac{1}{\omega}(\vec k\times \vec E) =\frac{1}{10c}\cdot 10(6\hat k-8\hat i)\cos(6x+8z-10ct)$$ so $$\boxed{\vec B=\frac{1}{c}(6\hat k-8\hat i)\cos(6x+8z-10ct)}$$$
Match with options

This matches Option B:

$\frac{1}{c}(6\hat k-8\hat i)\cos(6x+8z-10ct)$
Verification with stored answer

Stored correct answer: B

Our derived answer: B

Hence they agree.

View written solutionFree

Correct answer: B

Given electric field at $t=0$

$\vec E(x,z,0)=10\,\hat j\cos(6x+8z)$

For a plane electromagnetic wave, the general form is

$\vec E=\vec E_0\cos(\vec k\cdot \vec r-\omega t)$

or equivalently with a plus sign depending on propagation direction.
Identify the wave vector

From

$\cos(6x+8z)$

we get

$\vec k=6\hat i+8\hat k$

and its magnitude is

$|\vec k|=\sqrt{6^2+8^2}=10$

Hence

$\omega=c|\vec k|=10c$

So the time-dependent phase must be

$6x+8z-10ct$

for propagation along $+\vec k$ .
Direction of magnetic field

For an electromagnetic wave in free space,

$\vec B=\frac{1}{\omega}\,\vec k\times \vec E$

or equivalently,

$\vec B=\frac{1}{c}\,\hat n\times \vec E$

where $\hat n=\dfrac{\vec k}{|\vec k|}$ is the propagation direction.

Here,
$\qquad \vec k=6\hat i+8\hat k$$ Compute cross product: $$\vec k\times \vec E_0=(6\hat i+8\hat k)\times 10\hat j$$ $$=10\left[6(\hat i\times \hat j)+8(\hat k\times \hat j)\right]$$ Using $$\hat i\times \hat j=\hat k, \qquad \hat k\times \hat j=-\hat i$$ we get $$\vec k\times \vec E_0=10(6\hat k-8\hat i)$$ Therefore, $$\vec B=\frac{1}{\omega}(\vec k\times \vec E) =\frac{1}{10c}\cdot 10(6\hat k-8\hat i)\cos(6x+8z-10ct)$$ so $$\boxed{\vec B=\frac{1}{c}(6\hat k-8\hat i)\cos(6x+8z-10ct)}$$$
Match with options

This matches Option B:

$\frac{1}{c}(6\hat k-8\hat i)\cos(6x+8z-10ct)$
Verification with stored answer

Stored correct answer: B

Our derived answer: B

Hence they agree.