24.2 Production of electromagnetic waves  (Page 3/14)

 Page 3 / 14

Relating $E$ -field and $B$ -field strengths

There is a relationship between the $E$ - and $B$ -field strengths in an electromagnetic wave. This can be understood by again considering the antenna just described. The stronger the $E$ -field created by a separation of charge, the greater the current and, hence, the greater the $B$ -field created.

Since current is directly proportional to voltage (Ohm’s law) and voltage is directly proportional to $E$ -field strength, the two should be directly proportional. It can be shown that the magnitudes of the fields do have a constant ratio, equal to the speed of light. That is,

$\frac{E}{B}=c$

is the ratio of $E$ -field strength to $B$ -field strength in any electromagnetic wave. This is true at all times and at all locations in space. A simple and elegant result.

Calculating $B$ -field strength in an electromagnetic wave

What is the maximum strength of the $B$ -field in an electromagnetic wave that has a maximum $E$ -field strength of $\text{1000 V/m}$ ?

Strategy

To find the $B$ -field strength, we rearrange the above equation to solve for $B$ , yielding

$B=\frac{E}{c}.$

Solution

We are given $E$ , and $c$ is the speed of light. Entering these into the expression for $B$ yields

$B=\frac{\text{1000 V/m}}{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=\text{3}\text{.}\text{33}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T},$

Where T stands for Tesla, a measure of magnetic field strength.

Discussion

The $B$ -field strength is less than a tenth of the Earth’s admittedly weak magnetic field. This means that a relatively strong electric field of 1000 V/m is accompanied by a relatively weak magnetic field. Note that as this wave spreads out, say with distance from an antenna, its field strengths become progressively weaker.

The result of this example is consistent with the statement made in the module Maxwell’s Equations: Electromagnetic Waves Predicted and Observed that changing electric fields create relatively weak magnetic fields. They can be detected in electromagnetic waves, however, by taking advantage of the phenomenon of resonance, as Hertz did. A system with the same natural frequency as the electromagnetic wave can be made to oscillate. All radio and TV receivers use this principle to pick up and then amplify weak electromagnetic waves, while rejecting all others not at their resonant frequency.

Take-home experiment: antennas

For your TV or radio at home, identify the antenna, and sketch its shape. If you don’t have cable, you might have an outdoor or indoor TV antenna. Estimate its size. If the TV signal is between 60 and 216 MHz for basic channels, then what is the wavelength of those EM waves?

Try tuning the radio and note the small range of frequencies at which a reasonable signal for that station is received. (This is easier with digital readout.) If you have a car with a radio and extendable antenna, note the quality of reception as the length of the antenna is changed.

Phet explorations: radio waves and electromagnetic fields

Broadcast radio waves from KPhET. Wiggle the transmitter electron manually or have it oscillate automatically. Display the field as a curve or vectors. The strip chart shows the electron positions at the transmitter and at the receiver.

Section summary

• Electromagnetic waves are created by oscillating charges (which radiate whenever accelerated) and have the same frequency as the oscillation.
• Since the electric and magnetic fields in most electromagnetic waves are perpendicular to the direction in which the wave moves, it is ordinarily a transverse wave.
• The strengths of the electric and magnetic parts of the wave are related by
$\frac{E}{B}=\text{c},$

which implies that the magnetic field $B$ is very weak relative to the electric field $E$ .

Conceptual questions

The direction of the electric field shown in each part of [link] is that produced by the charge distribution in the wire. Justify the direction shown in each part, using the Coulomb force law and the definition of $\mathbf{E}=\mathbf{F}/q$ , where $q$ is a positive test charge.

Is the direction of the magnetic field shown in [link] (a) consistent with the right-hand rule for current (RHR-2) in the direction shown in the figure?

Why is the direction of the current shown in each part of [link] opposite to the electric field produced by the wire’s charge separation?

In which situation shown in [link] will the electromagnetic wave be more successful in inducing a current in the wire? Explain.

In which situation shown in [link] will the electromagnetic wave be more successful in inducing a current in the loop? Explain.

Should the straight wire antenna of a radio be vertical or horizontal to best receive radio waves broadcast by a vertical transmitter antenna? How should a loop antenna be aligned to best receive the signals? (Note that the direction of the loop that produces the best reception can be used to determine the location of the source. It is used for that purpose in tracking tagged animals in nature studies, for example.)

Under what conditions might wires in a DC circuit emit electromagnetic waves?

Give an example of interference of electromagnetic waves.

[link] shows the interference pattern of two radio antennas broadcasting the same signal. Explain how this is analogous to the interference pattern for sound produced by two speakers. Could this be used to make a directional antenna system that broadcasts preferentially in certain directions? Explain.

Problems&Exercises

What is the maximum electric field strength in an electromagnetic wave that has a maximum magnetic field strength of $5\text{.}\text{00}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{T}$ (about 10 times the Earth’s)?

150 kV/m

The maximum magnetic field strength of an electromagnetic field is $5×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}$ . Calculate the maximum electric field strength if the wave is traveling in a medium in which the speed of the wave is $\text{0.75}c$ .

Verify the units obtained for magnetic field strength $B$ in [link] (using the equation $B=\frac{E}{c}$ ) are in fact teslas (T).

how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
haider
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haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
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