Question

The magnetic field of a plane electromagnetic wave is given by:
$\vec B=2×10^{−8}sin⁡(0.5×10^3x+1.5×10^{11}t)\hat jT$.
The amplitude of the electric field would be

• A. 6 Vm–1 along x-axis
• B. 3 Vm–1 along z-axis
• C. 6 Vm–1 along z-axis
• D. 2 × 10–8 Vm–1 along z-axis

Answer 1

Answer: (C)

6 Vm–1 along z-axis

Answer 2

Speed of light,
$c =\frac wk$
$c=\frac {1.5×10^{11}}{0.5×10^3}$
$c=3×10^8\ m/sec$
So, $E_0=B_0c$
$E_0=2×10^{−8}×3×10^8$
$E_0=6\ V/m$
And the direction will be along z-axis.

So, the correct option is (C): 6 Vm–1 along z-axis