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

Options

A. `(1)/(c) ( 6 hat(k) + 8 hat(i) ) cos [ ( 6 x - 8 z + 10 ct)]`
B. `(1)/(c) ( 6 hat(k) - 8 hat(i) ) cos [ ( 6 x + 8 z - 10 ct)]`
C. `(1)/(c) ( 6 hat(k) + 8 hat(i) ) cos [ ( 6 x - 8 z + 10 ct)]`
D. `(1)/(c) ( 6 hat(k) - 8 hat(i) ) cos [ ( 6 x + 8 z + 10 ct)]`

Answer

B

Explanation

`vec (E) = E_(0) cos ( omega t - vec (k) - vec (r))`

`:. ` At any time `t = vec (k) = 6 hat(i) + 8 hat(k) `
`vec (E) = 10 hat(j) cos (6 x + 8 z - 10 ct) = | vec (k)| = 10`
` B_(0) = (E_(9))/( c) = (10)/(c) = lambda = (2 pi)/( k) = (2 pi)/(10) = (pi)/(5)`
Also ` vec (E) xx vec (B) = vec (c) = :. omega = 2 pi f = ( 2 pi c)/( lambda) `
` |{:(hat(i) ,hat(j),hat(k)),(0,1,0),(B_(x),B_(y),B_(2)):}| = ( 6 hat(i) + 8 hat(k))/( 10) omega = (2 pi xx c)/(((pi)/(5))) = 10 c`
`:. i [ B_(z) - 0] + hat(j)_ (0) + hat(k) ( - B_(x)) = 0.6 hat(i) + 0.8 hat(k)`
Comparing coefficient both sides
`:. B_(z) = 0.6 , B_(x) = - 0.8`
`| bar(B)_(0) | = (10)/(c) ( - 0. 8 hat(i) + 0.6 hat(k)) = ((-8 i + 6 hat(k))/(c))`
So magnetic field
`B = (1)/(c) ( - 8 hat(i) + 6 hat(k)) cos ( 6 x + 8 z - 10 ct )` Tesla