Question

An isotropic point source emits light at wavelength$\mathbf{500}\mathbf{}\mathbf{\text{nm}}$, at the rate of$\mathbf{200}\mathbf{}\mathbf{\text{W}}$. A light detector is positioned$\mathbf{400}\mathbf{}\mathbf{\text{m}}$from the source. What is the maximum rate$\mathbf{\partial }\mathit{B}\mathbf{/}\mathbf{\partial }\mathit{t}\mathbf{}$ at which the magnetic component of the light changes with time at the detector’s location?

Short answer

The maximum rate of $\partial B/\partial t$ is $3.44\times {10}^6\text{T/s}$.

Step-by-step solution

Listing the given quantities

Wavelength is, $\lambda =500\text{nm}$.

Power rate is, $P=200\mathrm{W}$.

Distance is, $r=400\mathrm{m}$.

Understanding the concepts of the partial derivative of the magnetic field

Taking the partial derivative ofthemagnetic fieldrelation, we get the maximum value of the rate of change of the magnetic field. Finding the value of angular frequency and magnitude of magnetic field component and substituting in partial derivative, we can find the value of the maximum rate of change of magnetic field.

Determination of maximum rate

The magnetic field is a sinusoidal function of position x and time tand is given by,

$B=B_m\sin \left(kx-\omega t\right)$

Here $\omega$ is the angular frequency and $B_m$is the amplitude.

Taking the partial derivative with respect to time, we get,

$\frac{\partial B}{\partial t}=-\omega B_m\cos \left(kx-\omega t\right)$

This shows that the maximum value of $\frac{\partial B}{\partial t}$is $\omega B_m$because the maximum value of the cosine function is 1.

${\left(\frac{\partial B}{\partial t}\right)}_{max}=\omega B_m$ (1)

Now we know the wave speedc and amplitudes of electric and magnetic fields are related as,

$\begin{array}{rcl}c& =& \frac{E_m}{B_m}\\ B_m& =& \frac{E_m}{c}\end{array}$

The magnetic field in terms of intensity is,

$B_m=\frac{\sqrt{2c{\mu }_{0}I}}{c}$

$B_m=\frac{\sqrt{2c{\mu }_{0}I}}{c}$

The intensity and the power are related as,

$I=\frac{P}{4\pi r^2}$

Substituting this value in $B_m$, we get,

$\begin{array}{rcl}{B}_m& =& \frac{\sqrt{2c{\mu }_0\frac{P}{4\pi r^2}}}{c}\\ B_m& =& \sqrt{\frac{{\mu }_{0}P}{2\pi c}}\frac{1}{r}\end{array}$ (2)

Now we know that,

$\begin{array}{rcl}\omega & =& 2\pi f\\ & =& \frac{2\pi c}{\lambda }\end{array}$ (3)

Substituting values from equations (2) and (3) in equation (1), we get,

${\left(\frac{\partial B}{\partial t}\right)}_{\max }=\sqrt{\frac{{\mu }_{0}P}{\pi c}}\frac{2\pi c}{\lambda r}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{\left(\frac{\partial B}{\partial t}\right)}_{\max }& =& \sqrt{\frac{\left(4\pi \times {10}^{-7}\text{T}·\frac{\text{m}}{\text{A}}\right)\left(200\text{W}\right)}{\pi \left(3\times {10}^8\frac{\text{m}}{\text{s}}\right)}}\times \frac{2\pi \left(3\times {10}^8\text{m/s}\right)}{\left(500\times {10}^{-9}\text{m}\right)\left(400\text{m}\right)}\\ & =& 3.44\times {10}^6\text{T/s}\end{array}$

Thus, the maximum rate of $\partial B/\partial t$is $3.44\times {10}^6\text{T/s}$.