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A parallel plate capacitor with circular plates of radius R = 16.0 cm and plate separation d = 9.00 mm is being charged at the rate of 8.00 C/s.

What is the displacement current through a circular loop of radius r = 21.00 cm centered on the axis of the capacitor? 8.00 You are correct.

What is the displacement current through a circular loop of radius r = 3.00 cm centered on the axis of the capacitor?

What is the magnitude of the magnetic field between the capacitor plates at a radius r = 3.00 cm from the axis of the capacitor?

Sagot :

Part a: Displacement current: I(d1) = 8 ampere.

Part b: displacement current: I(d2) = 2.812 x 10⁻¹ A.

Part c: magnetic field at a r = 3 cm radial distance:  1.875 x 10⁻⁶ T.

Explain the term Displacement Current?

  • The electric field that exists between both the capacitor plates stores energy in capacitors.
  • A current known as displacement current forms between the capacitor plates if the electric field between them is continuously changing.

Part a: Displacement current within the 21 cm-radius circular loop.

Since the capacitance plate has a 16 cm radius, the current can only flow in a loop of this size, or the total displacement current flows within a loop with a 21 cm radius.

Displacement current;

I(d1) = ∈₀.dФ(e)/dt

I(d1) = ∈₀ x A x (Qr.d/∈₀.A.d)

I(d1) = Qr

I(d1) = 8 C/s

I(d1) = 8 ampere of Displacement current.

Part b: displacement current flowing through a 3 cm-diameter circular loop

Density of the displacement current;

Jd = I(d1)/A

Jd = 8/(8.042 x 10⁻²)

Jd = 99.478 A/m²

For the radius r, area of loop

A1 = πr²

A1 = 3.14 x (0.03)²

A1 = 2.827 x 10⁻³ m²

Thus, displacement current;

I(d2) = Jd x A1

I(d2) = 99.478 x 2.827 x 10⁻³

I(d2) = 2.812 x 10⁻¹ A.

Part c: magnetic field at a r = 3 cm radial distance.

Field of magnetism:

B = μ₀.I(d2) / 2πr

B = 4π x 10⁻⁷ x 2.812 x 10⁻¹ / 2π x 0.03

B = 1.875 x 10⁻⁶ T

Thus, magnetic field at a r = 3 cm radial distance is 1.875 x 10⁻⁶ T.

To know more about the Displacement Current, here

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