Question

2. A plane electromagnetic wave travelling along X-axis has a wavelength 10.0 mm. The electric
field points along Y-direction and has peak value of 30 V/m. Then the magnetic field in terms
of x in metre and t in second may be expressed as
[NCERT Exemplar|​

Answer 1

Answer:

See below.

Explanation:

We are given that the EM wave is travelling along the X-axis. The electric field is along the Y-axis which implies that the magnetic field is along the Z-axis.

The peak value of the electric field,

$E_o = 30 \ V/m$

The peak value of the magnetic field will be,

$B_o = \frac{E_o}{c} = \frac{30}{3 \times 10^{8}} = 10^{-7} \ T$

We are given the wavelength of the EM wave. Calculating the value of k,

$k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{ 10 \times 10^{-3}} = 2 \pi \times 10^2 = 628 \ m^{-1}$

We know,

$v=\dfrac{\omega }{k}$

Therefore, the angular frequency of the EM wave will be,

$\omega = c \cdot k = ( 3 \times 10^8) \times (2 \pi \times 10^2 ) = 6 \pi \times 10^{10} = 1.88 \times 10^9$

The component of the magnetic field will be given by,

$B_z = B_o \ sin( \omega t - kx )$

Substituting the values,

$B_z = ( 10^{-7}) \ sin[ (1.88 \times 10^9 )t - (628)x ]$