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marysya [2.9K]
3 years ago
13

A parallel-plate capacitor with circular plates is charged by a constant current III. The radius aaa of the plates is much large

r than the distance ddd between them, so fringing effects are negligible. Calculate B(r)B(r)B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates.
Physics
1 answer:
shutvik [7]3 years ago
4 0

Answer:

Please refer to the explanation below.

Explanation:

Let

Radius of parallel plate capacitor = r

Distance between parallel plates of the capacitor = R

Area of parallel plate capacitor = Ar

Area of Distance between parallel plates of the capacitor = AR

Constant Current = I₀

Permeability of free space = μ₀

Induced magnetic field = B

From the Ampere's Maxwell law the magnetic field is given by

∫B.ds = μ₀I₀

∫B.ds = μ₀I₀Ar/AR

When the radius of the plates is much less than the distance between the parallel plates (r < R)

∫B.ds = μ₀I₀πr²/πR²

∫B.ds = μ₀I₀r²/R²

s = 2πr

B.2πr = μ₀I₀r²/R²

B = μ₀I₀r²/2πrR²

B = μ₀I₀r/2πR²

When the radius of the plates is much larger than the distance between the parallel plates (r > R)

∫B.ds = μ₀I₀

B.2πr = μ₀I₀

B = μ₀I₀/2πr

Bonus:

The magnetic field will be maximum when the radius of the parallel plate capacitor is equal to he distance between the parallel plate capacitor that is r = R

B = μ₀I₀/2πR

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Answer:

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Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

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b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

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Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

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Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

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The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

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The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

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Heat required to heat up to melting point,

Q = Q_1+Q_2

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