Question

The sunlight reaching the earth has a maximum electric field of $810\;V/m$. What is the maximum magnetic field in this light?

Answer 1

The sunlight is a form of electromagnetic radiation and hence the concepts of electromagnetic and hence the concept of electromagnetic radiation is applied.The formula relating the electric field and the magnetic field is applied in order to determine the magnetic field of the light. The values of constants used in the formula must be applied to determine the answer to the question.

Complete step by step answer:
The above problem revolves around the concept of electromagnetic radiation and the relation between the electric and magnetic fields. In order to find the magnetic field given the electric field of light we first need to look into the concept of electromagnetic radiation.Maxwell put forward a theory that when there is a change in the electric or field then the other field, that is, the magnetic field was produced and this observation applied for experiments where a change in magnetic field resulted in the production of electric field as well. Hence, it was said that there was a relation between these two fields as both of these fields acted as the sources of production of one another.

Let us now define electromagnetic radiation. An electromagnetic wave is said to be the wave that is radiated by an accelerated charge which propagates in space as coupled electric and magnetic fields that are produced mutually perpendicular to each other and to the direction of propagation of the wave.

Maxwell proposed that only an accelerating charge particle can produce electromagnetic properties as the oscillating charge particle provides an electric field around it and similarly produces a magnetic field along with it since the particle is in motion. Electromagnetic radiation is said to be a spectrum that includes various types of sections based on their differing wavelength. These include radio waves, microwaves, infrared radiation, visible light, ultra violet, X-rays and gamma rays which all exhibit electromagnetic properties.

The most common type of electromagnetic radiation is visible light and sunlight is an example of visible light as visible light is one that can be seen to the naked eye. The other forms of EM waves are not visible to the eye. Here, we are considering sunlight which exhibits EM wave properties as given in the question and hence sunlight will have both electric field and magnetic field that is produced and varies with each other.

We now need to determine a relation between the electric and magnetic fields that are present in sunlight. The ratio of amplitudes, that is, the maximum values of electric and magnetic fields is equivalent to the speed of light. The value for speed of light can be directly applied and the given electric field value can be applied to directly find out the value for magnetic field. The equation relating the amplitudes of the electric and magnetic fields in terms of the speed of light is given as:
$\dfrac{{{E_0}}}{{{B_0}}} = c$
We cross multiply the terms to get:
${E_0} = {B_0} \times c$
We make magnetic fields the subject since the maximum magnetic field is asked to be found out. Hence, we get:
$\Rightarrow {B_0} = \dfrac{{{E_0}}}{c}$ --------($1$)

Given, ${E_0} = 810\;V/m$. We know that the speed of light is constant and has a default value, that is, $c = 3 \times {10^8}m/s$. Hence we substitute these values to get:
$\Rightarrow {B_0} = \dfrac{{810}}{{3 \times {{10}^8}}}$
$\Rightarrow {B_0} = \dfrac{{810}}{3} \times {10^{ - 8}}$
$\Rightarrow {B_0} = 270 \times {10^{ - 8}}$
The unit of magnetic field is given as tesla denoted by ‘T’. We shift the decimal places to the left by two places to get:
$\Rightarrow {B_0} = 2.7 \times {10^{ - 6}}T$
We know that,
$1\mu T = {10^{ - 6}}T$
Hence,
$\therefore {B_0} = 2.7\mu T$
This is the maximum magnetic field of sunlight reaching the Earth.

Note: Maxwell also gave an equation relating the amplitudes of the electric and magnetic fields in terms of the permittivity of free space and permeability values. An alternative way to solve this problem is by determining an equation between electric and magnetic fields in terms of them and substituting their default values to determine the magnetic field.