The electric field in a plane electromagnetic wave is given by

E_z=60\cos (5x+1.5\times {10}^{9}t)\frac{V}{m}

Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

We are given the electric field of a plane electromagnetic wave

$E_z=60\cos (5x+1.5\times {10}^{9}t)Vm⁻¹$

In a uniform plane wave propagating in the + x‑direction, the fields are mutually perpendicular and satisfy the relation

$B=\frac{E}{c}$

where c = 3 × 10⁸ m s⁻¹ is the speed of light in vacuum. The direction of B is given by \(\hat{k}\times\hat{E}\). Since the wave travels along + x, \(\hat{k} = \hat{x}\) and \(\hat{E} = \hat{z}\); therefore \(\hat{x}\times\hat{z} = \hat{y}\). Hence the magnetic field points in the y‑direction:

$B_y=\frac{E_z}{3}$

Substituting the given E‑field amplitude 60 V m⁻¹:

$B_y=\frac{60}{3}\cos (5x+1.5\times {10}^{9}t)T$

Evaluating the fraction:

$\frac{60}{3}=2\times {10}^{-7}$

Thus the magnetic field expression is

$B_y=2\times {10}^{-7}\cos (5x+1.5\times {10}^{9}t)T$

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