Physics, 08.04.2020 01:28 Arinola

5.0-A current. (a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center. (b) Along the axis, at what distance from the center of the coil is the field magnitude 1 8 as great as it is at the center

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Answer from: jrich5111

Here is the full question

A coil consisting of 100 circular loops with radius 0.6 m carries a 5.0-A current.

(a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center.

(b) Along the axis, at what distance from the center of the coil is the field magnitude 1/8 as great as it is at the center

a) B = 0.000113 T

b) The distance (x) from the center of the coil = 1.038 m

Explanation:

The formula for the magnetic field on the line of axis of the circular loop can be expressed as:

B = $\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}$

where;

N = number of turns = 100

$\mu =$ $4 \pi * 10^{-7} \ N/m$

I = current flowing through the coil = 5.0- A

r = radius of the circular loop = 0.6 m

x = distance from the center of the coil

So;

$B = \frac{4 \pi *10^{-7}N/m (100)(5A)(0.6\ m^2)}{2((0.6 \m )^2+(0.8m)^2)^{3/2}}$

B = 0.000113 T

Also; to determine the distance (x) from the center of the coil; we have the following:

We know that the magnetic field at the center of the coil can be expressed as:

$B = \frac{\mu_o NI}{2 r}$

Now; given that the field magnitude is 1/8 as great as it is at the center; Then ;

$B' = \frac{1}{8} B$

∴

$\frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}$ = $\frac{1}{8} ( \frac{\mu_o NI}{2 r})$

$\frac{r^2}{(r^2+x^2)^{3/2}}= (\frac{1}{8}) r$

$8r^3 = (r^2+x^2)^{3/2}$

(8r^3)^2 = (r^2+x^2)^3

(64r^6) = (r^2+x^2)^3

$(r^2+x^2) = \sqrt[3]{64r^6}$

(r^2+x^2) = 4r^2

x^2 = 4r^2- r^2

$x^2 = 3 \ r^2$

$x= \sqrt {3 r^2}$

x = 1.73 (r)

x = 1.73 (0.6)

$x = 1.038 \ m$

Therefore, the distance from the center of the coil is = 1.038 m

Answer from: Quest

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Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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5.0-A current. (a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the ce...