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Nitella [24]
3 years ago
10

An infinitely long wire carrying a 25-A current in a the positive x-direction is places along the xaxis in the vicinity of a 20-

turn circular loop located in the x-y plane. If the magnetic field at the center of the loop is zero, what is the direction and magnitude of the current flowing in the loop
Physics
1 answer:
Masteriza [31]3 years ago
8 0

Answer:

0.2 A, clockwise direction

Explanation:

We are given that

Current,I=25 A

Number of turns=n=20

Magnetic field at center of r the loop=B=0

d=2 m

We have to find the direction and magnitude of current flowing in the loop.

Magnetic field due to current I

B=\frac{\mu_0 I}{2\pi d}\hat{z}

Magnetic field due to I'

B'=-\frac{\mu_0 nI''}{2a}\hat{z}

a=1 m

Net magnetic is zero

Therefore, B+B'=0

\frac{\mu_0 I}{2\pi d}\hat{z}-\frac{\mu_0 nI''}{2a}\hat{z}=0

\frac{\mu_0 I}{2\pi d}\hat{z}=\frac{\mu_0 nI''}{2a}\hat{z}

\frac{I}{\pi d}=\frac{I'n}{a}

I'=\frac{aI}{\pi d n}=\frac{25\times 1}{3.14\times 2\times 20}=0.2 A

Where \pi=3.14

Direction: Clockwise

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Explanation:

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Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

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M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

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<u />

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