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4 years ago

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The magnetic field created by a large current passing through plasma (ionized gas) can force current-carrying particles together

. This pinch effect has been used in designing fusion reactors. It can be demonstrated by making an empty aluminum can carry a large current parallel to its axis. Let R represent the radius of the can and I the current, uniformly distributed over the can's curved wall. Determine the magnetic field just inside the wall.

1 answer:

RUDIKE [14]4 years ago

5 0

To develop this problem we must apply the concepts related to the Ampere Law. The line integral of B.ds around any closed path is equal to the current permeability constant, this current for the particular case passes through the 'internal' surface delimited by the closed path.

Ampere laws is defined as,

$\oint Bds=mu_0 I$

The radius of the circle is r, so if

r < R

And there is no current inside so ,

$\oint Bds=mu_0 I$

$B (2\pi R) = 0$

Therefore the magnetic field inside the wall is Zero.

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Answer:

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let's calculate

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Julli [10]

Explanation:

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Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

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$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

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$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

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$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

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Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

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From equation (II)

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Complete Question

The complete question is shown on the first uploaded image

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