Electric and Magnetic Fields in Vacuum |
Born on June 13, 1831 in Edinburgh, Scotland. Maxwell died in Cambridge, England on November 5, 1879 |
James Clerk Maxwell |
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Laid the theoretical foundations for the theory of the electric field in his paper on electricity and magnetism where he established the complete electromagnetic theory of light based on Michael Faraday's ideas. This showed that oscillating charges produce waves in an electromagnetic field (1873) |
Electric field produced by changing magnetic field |
Magnetic field produced by changing electric field |
\[\oint \vec E \cdot d \vec S = - \frac{d\Phi_B}{dt}\]
\[\oint \vec B \cdot d \vec S =𝜇_0𝜀_0 \frac{d\Phi_E}{dt}\]Note that Maxwell's equation is derived from Faraday's laws of electromagnetic induction
It states that "when there is a circular charged capacitor through a time-varying electric field
an induced magnetic field is generated."
With this step, the laws of forces in nature were unified
And through Ampere's law and the generation of a magnetic field around a wire \[\oint \vec B \cdot d \vec S =𝜇_0 . I_{enc}\]Maxwell noticed a deficiency in the law and reached the following result by combining Maxwell's and Ampere's laws \[\oint \vec B \cdot d \vec S =𝜇_0𝜀_0 \frac{d\Phi_E}{dt}+𝜇_0 . I_{enc}\]In the case of a steady current in the conductor, the law is limited to Ampere's law
And in the case of a varying field without current in the conductor, we limit it to Maxwell's law
From Maxwell-Ampere's equation we call\[ I_d=𝜀_0 \frac{d\Phi_E}{dt}\]Displacement current
So Maxwell-Ampere's equation becomes \[\oint \vec B \cdot d \vec S =𝜇_0(I_d+I_{enc})\]
Solved Example
An electric field was directed perpendicular to a circular surface of radius \[8\;cm\] If the field increases at a time rate \[15\frac {v}{m.s}\] then the magnitude of the magnetic field at a distance from the center of the circular area by \[20\;cm\] SolutionFirst we calculate the displacement current\[ I_d=𝜀_0\frac{d\Phi_E}{dt}\]\[ I_d=𝜀_0.A\frac{∆𝐸}{∆t}\]\[ I_d=8.85×10^{-12}×𝜋(0.08)^2×15=2.67×10^{−12} A\]
Then we calculate the field \[B=\frac{𝜇_0.𝑖}{2𝜋 𝑟}\]\[B=\frac{4𝜋×2.67×10^{−12}}{2𝜋×0.2}=2.67×10^{−18}T\] Information for knowledge: Measuring the speed of light
Danish astronomer Rømer (1644-1710) noticed
while observing Jupiter and its moon Io
Rømer noticed that when Earth was far from Jupiter
Io began to hide later than when it was close
The speed of light was calculated assuming the delay time
This corresponds to the time it takes light to cross Earth's orbit
The calculated speed of light at that time was an inaccurate value 30% less than what is currently known
The speed of light is currently estimated at 299,792,458 meters per second
C= 3.0 × 108m/s
Visible light is one component of electromagnetic waves
All electromagnetic waves travel
in vacuum at the same speed \[C= 3 × 10^8 m/s \] \[C=v=\frac{X}{t}=\frac{\lambda }{T}=\lambda f \]
c =v |
X |
t |
λ |
T |
\[f \] |
Speed of light |
Distance traveled |
Time |
Wavelength |
Periodic time |
Frequency |
The wavelength of blue light is \[λ=5×10^{-7} m\]
Then the frequency of the light is
Click here to show solution
Choose the correct answer
What is the frequency of an electromagnetic wave with wavelength
3.2 × 10-7m
Click here to show solutionChoose the correct answer
Types of Electromagnetic Waves
The main source of electromagnetic waves is the sun
Electromagnetic waves are divided into seven types
Infrared waves |
Microwaves |
Radio waves |
Used in night vision cameras - remote control devices |
Used in microwave ovens - radars |
Used in radio and television broadcasting devices |
X-rays |
Ultraviolet waves |
Visible light waves |
Used in medical imaging - treating some cancer tumors |
Used in sterilizing drinking water - sterilizing medical tools |
Used in homes, microscopes and seeing objects |
Gamma rays |
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Used in treating some cancer tumors |
Transmission of Electromagnetic Waves
Waves are transmitted via an antenna, and the potential difference between the ends of the antenna accelerates the movement of charges which in turn create a changing magnetic field due to their motion that generates a changing electric field and so on, propagating in vacuum at the speed of light
The changing electric field produces a changing magnetic field which in turn produces a changing electric field and so on
Speed of wave propagation in a material medium
Less than the speed of the wave in vacuum Given by the relation \[ v=\frac{{{C}}}{{{\sqrt K }}}\] (K) where
is the dielectric constant which varies according to the medium and equals \[ \sqrt K = n \] \[(n )\] is the refractive index of the medium
If the dielectric constant of glass is \[ K=2.28 \]
Choose the correct answer
Then the speed of light in glass equals
Click here to show solution
Carrier waves: Radio broadcasts at specific wavelengths, each radio has a carrier wave whose frequency or amplitude is modified by the transmitter
The transmitter consists of three devices
Oscillator: generates the carrier wave
Modulator: modifies the frequency or amplitude of the carrier wave that carries sound and image data
Amplifier: increases the signal voltage
Frequency tuning: done via an LC circuit
In this simulation, an oscillating circuit consisting of a capacitor and an inductor
The capacitor is charged by a battery then disconnected and the charged capacitor is connected to an inductor and observe how the circuit works
Thus creating a changing electric field and a changing magnetic field with a specific frequency
Polarization
Light is the interaction of electric and magnetic fields traveling through space. The electric and magnetic vibrations of a light wave occur perpendicular to each other. The electric field moves in one direction and the magnetic field in another direction perpendicular to each other. So, we have one plane occupied by an electric field, another plane occupied by a perpendicular magnetic field, and the wave propagation direction perpendicular to both. These electric and magnetic vibrations can occur in many planes. A light wave vibrating in more than one plane is called unpolarized light. Light emitted from the sun, a lamp, or a tube are sources of unpolarized light.
Light polarization allows the passage of light's component fields in one direction only or allows the passage of one field more than another If we have light of intensity \[I_0\] and place a vertical filter in front of it (the filter's axis parallel to one of the fields), it allows one field to pass and blocks the other, and the polarized light intensity becomes \[I=\frac{I_0}{2}\]
And if there's an angle between the filter's axis and the light's direction (neither parallel nor perpendicular), the filtered light intensity becomes \[I=I_0 Cos^2(𝜃)\]
Solved Example
What angle is required between the direction of polarized light and the polarization filter's axis to reduce its intensity by
60%
Solution
When intensity decreases by60%
the ratio between the transmitted intensity to the original intensity equals
40% \[\frac{I}{I_0}=0.4\]\[I=I_0 Cos^2(𝜃)\]\[\frac{I}{I_0}= Cos^2(𝜃)\]\[\sqrt\frac{I}{I_0}=Cos(𝜃)\]\[𝜃=Cos^{-1}\sqrt {O.4}=50.7^0\]
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