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Ratling [72]
3 years ago
7

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

b. What is the E-field amplitude? c. What is the intensity of this wave? What pressure does such wave exert on a totally-absorbing surface?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

Part a)

\lambda = 300 m

Part b)

E = 2.7 N/C

Part c)

I = 9.68 \times 10^{-3} W/m^2

P = 3.22 \times 10^{-11} N/m^2

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

\lambda = \frac{c}{f}

\lambda = \frac{3\times 10^8}{1\times 10^6}

\lambda = 300 m

Part b)

As we know the relation between electric field and magnetic field

E = Bc

E = (9 \times 10^{-9})(3\times 10^8)

E = 2.7 N/C

Part c)

Intensity of wave is given as

I = \frac{1}{2}\epsilon_0E^2c

I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)

I = 9.68 \times 10^{-3} W/m^2

Pressure is defined as ratio of intensity and speed

P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}

P = 3.22 \times 10^{-11} N/m^2

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Physics B 2020 Unit 3 Test
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Answer:

1)

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F=qvB sin \theta

where here:

For the proton in this problem:

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B = 19 T is the magnetic field

\theta=65^{\circ} is the angle between the directions of v and B

So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

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3)

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It can be applied as follows:

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4)

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r=\frac{mv}{qB}

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v=2155 m/s is the speed of the alpha particle

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r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

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f=\frac{qB}{2\pi m}

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7)

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8)

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So the flux is

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See the last 7 answers in the attached document.

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3 years ago
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Answer:

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