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3 years ago

Which part of the electromagnetic spectrum is divided into seven ranges of wavelengths and is the part visible to the human eye

2 answers:

6 0

In one of the most amazing coincidences in all of science,
the part of the electromagnetic spectrum that's visible to the
human eye is called "visible light".

Visible light is not 'divided' into anything. We mention the names
to seven of the colors in visible light. But all of the thousands of
OTHER colors that we can see are in there too, even though we
don't bother to list their names when we buzz through the rainbow
in the third grade.

Mice21 [21]3 years ago

6 0

Answer: B. visible light

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.

Electromagnetic spectrum consists of electromagnetic waves called as radio waves, microwaves, infrared, Visible, ultraviolet , X rays and gamma rays in order of increasing frequency and decreasing wavelength.

Visible light is composed of seven colors as violet, indigo , blue , green, yellow, orange and red light which are arranged in order of increasing wavelengths which is visible to naked eye.

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

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Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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2 years ago

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