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Aleksandr-060686 [28]
3 years ago
13

A bar magnet has five lines of force cutting one square centimeter (1 cm2). Assuming that each line of force represents 30 Wb of

flux, the flux density for the magnet is
Physics
1 answer:
user100 [1]3 years ago
4 0

Answer:

The magnetic flux density is 3 x 10⁵ Wb/m²

Explanation:

Given;

magnetic flux of each line of force, Ф = 30 Wb

area of the bar magnet, A = 1 cm² = 1 x 10⁻⁴ m²

The flux density of the magnet is the measure of the concentration of the magnetic flux per unit area. The unit is Wb/m² or Tesla.

B = Ф / A

B = 30 / (1 x 10⁻⁴ )

B = 3 x 10⁵ Wb/m²

Therefore, the magnetic flux density is 3 x 10⁵ Wb/m²

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To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

V = \sqrt{\frac{T}{\mu}}

Our values are given as:

f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w

Replacing we have that the velocity is

V = \sqrt{\frac{T}{\mu}}

V = \sqrt{\frac{65}{0.230}}

V = 16.81m/s

From the theory of wave propagation the average power wave is given as

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A = Amplitude

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A^2 = \frac{2P}{\mu \omega^2 V}

A^2 = \frac{2P}{\mu (2\pi f)^2 V}

Replacing,

A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}

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8 0
3 years ago
What amount of energy is needed for an electron to jump from n = 1 to n = 4?
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Answer:

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We need to find the energy for an electron to jump from n = 1 to n = 4.

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The difference in energy for n = 1 to n = 4 is:

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So, the required energy is equal to 2.04\times 10^{-18}\ J.

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A laser emits a beam of light whose photons all have the same frequency. When the beam strikes the surface of a metal, photoelec
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the no. of ejected electrons per second will increase.

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A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, <u>the no. of ejected electrons per second will increase.</u>

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