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2 years ago

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Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has

a wavelength of 784 m?

1 answer:

dmitriy555 [2]2 years ago

7 0

Answer:

Explanation:

Given

Wavelength of radiation $\lambda =784\ m$

We know Energy of wave with wavelength $\lambda$ is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

c=velocity of light

$\lambda$ =wavelength of wave

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{784}$

$E=2.53\times 10^{-28}\ J$

Hence the energy of the wave with wavelength 784 m is $2.53\times 10^{-28}\ J$

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Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field

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Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is $\mu_0 = 1.32*10^{-6} \ Tm/A$

The percentage error is % error = 5.25%

Explanation:

From the question we are told that

The slope is $s= 3.9\ G/A$ , and

Generally $1 \ gauss = 10^{-4 } tesla$

So $s = 3.9 *10^ {-4} T /A$

From the relation given in the question

$\mu_0 = \frac{2R}{N} [\frac{B}{I} ]$

Where R is the radius of the coil $=\frac{Diameter \ of \ coil }{2} = 0.017m$

N is the number of loops of the coil = 10

Now from the question we are told that

$s = \frac{B}{I}$

substituting into the equation above

$\mu_0 = \frac{2R }{N} s$

Substituting values

$\mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}$

$= 1.326 *10^ {-6} \ Tm/A$

Generally the % error is mathematically represented as

% $error = \frac{Measured \ value - accepted \ value}{accepted \ value } *100$

Given that the accepted value is $\mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A$

Hence substituting values

% $error = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100$

$= 5.24$

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