Question

The electric field of a $450\mathrm{MHz}$ radio wave has a maximum rate of change of$4.5\times {10}^{11}(\mathrm{V}/\mathrm{m})/\mathrm{s}$ . What is the wave's magnetic field amplitude$?$

Short answer

The waves magnetic field amplitude is$0.523\mathrm{\mu T}$.

Step-by-step solution

Expression of electric field strength

Electric field strength is,

${\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{o}}\sin \left(2\mathrm{\pi }\left(\frac{\mathrm{x}}{\mathrm{\lambda }}-\mathrm{ft}\right)\right)$

$\frac{{\mathrm{dE}}_{\mathrm{y}}}{\mathrm{dt}}=-2{\mathrm{\pi fE}}_{\mathrm{o}}\cos \left(2\mathrm{\pi }\left(\frac{\mathrm{x}}{\mathrm{\lambda }}-\mathrm{ft}\right)\right)$

${\mathrm{E}}_{\mathrm{o}}=\frac{\left(\frac{{\mathrm{dE}}_{\mathrm{y}}}{\mathrm{dt}}\right)}{2\mathrm{\pi f}}$

Calculation of magnetic field amplitude

Given info:

Maximum rate of change is $4.5\times {10}^{11}(\mathrm{V}/\mathrm{m})/\mathrm{s}$.

Frequency is$450\mathrm{MHz}$.

Magnetic field amplitude is,

${\mathrm{B}}_{\mathrm{o}}=\frac{{\mathrm{E}}_{\mathrm{o}}}{\mathrm{c}}$

Substitute electric field strength expression,

We get,

${\mathrm{B}}_{\mathrm{o}}=\frac{\left(\frac{{\mathrm{dE}}_{\mathrm{y}}}{\mathrm{dt}}\right)}{2\mathrm{\pi fc}}$

Substitute all values,

${\mathrm{B}}_{\mathrm{o}}=\frac{4.5\times {10}^{11}(\mathrm{V}/\mathrm{m})/\mathrm{s}}{2\mathrm{\pi }\left(450\times {10}^6\mathrm{Hz}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}$

${\mathrm{B}}_{\mathrm{o}}=0.523\mathrm{\mu T}$