Question

A plane e.m wave of frequency $30MHz$ travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time $E=6V/m$ along $y$-direction. Its magnetic field component $B$ at this point would be

• A. $2\times {10}^{-8}T$ along $z$-direction
• B. $6\times {10}^{-8}T$ along $x$-direction
• C. $2\times {10}^{-8}T$ along $y$-direction
• D. $6\times {10}^{-8}T$ along $z$-direction

Answer 1

The correct option is A $2\times {10}^{-8}T$ along $z$-direction

Given data

frequency$=30MHz$

Electric field $\epsilon =\frac{6v}{m}\hat{J}$ (along + y axis)

Now we know that

$c=\frac{{\epsilon }_0}{B_0}$

c = speed of EM wave$=3\times {10}^{8}m/s$

${\epsilon }_0=$peak value of electric field

$B_0=$peak value of magnetic field

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

B = magnetic field

$3\times {10}^8=\frac{6}{B}$

$B=\frac{6}{3\times {10}^8}$ $B=2\times {10}^{-8}$ Tesla

Now we have to find direction of magnetic field

We know that wave propagates in the direction perpendicular to oscillating electric field and magnetic field.

$c=\overrightarrow{\epsilon }\times \overrightarrow{B}$ (only for direction)

c = along x axis

$\epsilon =$along y axis

So B should be along z axis

$B=2\times {10}^{-8}T$ along z axis