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sashaice [31]
3 years ago
11

What is the difference between mutual flux, leakage flux and magnetizing flux ​

Physics
1 answer:
Lemur [1.5K]3 years ago
3 0

In simple words, flux can be stated as the rate of flow of a fluid, radiant energy, or particles across a given area.

<u>Explanation:</u>

<u>Mutual Flux:</u>

  • The magnetic lines present in among two magnets or solenoid is mutual flux.
  • These are the lines in which the attraction and repulsion happens.
  • The SI unit of mutual flux is the Henry

<u>Leakage Flux:</u>

  • In simple words, it can be stated as  the magnetic flux which does not follow the specially designed way in a magnetic circuit.
  • Leakage flux in the induction motor takes spot due to current runs through the essence of the induction motor.
  • The SI unit of Leakage flux is the Weber

<u>Magnetizing flux</u>

  • Magnetic flux is an analysis of the entire magnetic field which moves in a given field
  • In simple words can be defined as the Magnetic flux is what generates the field around a magnetic material.
  • The SI unit of magnetic flux is the Weber
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Follows are the solution to this question:

Explanation:

In point a:

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X(t)=\int V_{x}(t)dt

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\to t=0\\\\ \to X(0)=2.3 \ m

\to X(0)=0+C\\\\ \to C=2.3\  m

\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\

        = \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\  = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m

In point b:

when t=2.2 \ s

the Particle velocity  (V)=2 \times 2.22 =9.68\  \frac{m}{s}

In point c:

Calculating the Particle acceleration:

\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2  =8.8 \ \frac{m}{s^2}

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Lets try to find how x^2+y^2+z^2-4x+4y=73 can be written in the form of (x-a)^2+(x-b)^2+(x-c)^2=R^2, since then its radius R and center of coordinates C=(a,b,c) would be easily recognizable.

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(x-2)^2-4=x^2-4x+4-4=x^2-4x, as we had originally with the x terms

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Which means:

x^2-4x+y^2+4y+z^2=(x-2)^2-4+(y+2)^2-4+z^2

Or, putting all together:

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