Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 35.0 mm and a plate separation of 4.1 mm.
Suppose also that a sinusoidal potential difference with a maximum value of 160 V and a frequency of 60 Hz is applied across the plates; that isV=(160.0 V)sin((2.*p)*(60 Hz * t)).Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.Find B(r = 17.5 mm).Find B(r = 70.0 mm).Find B(r = 105.0 mm).

Answer 1

Answer:

β max = 2.86 x 10 ⁻¹² T

β max = 1.432 x 10 ⁻¹² T

β max = 5.726 x 10 ⁻¹² T

β max = 8.589 x 10 ⁻¹² T

Explanation:

Given:

R = 35.0 mm , d = 4.1 mm , f = 60 Hz , V = 160v

And knowing

μ₀ = 4 π x 10 ⁻⁷ T * m / A ,  ε₀ = 8.85 x 10⁻¹² C² / N * m²

To find β max can use the equation

β max = [ π * f * μ₀ * ε₀ * r * V ] / d

r = R

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.035 m * 160 ] / (4.1 x 10⁻³ )

β max = 2.86 x 10 ⁻¹² T

r = 17.5 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.0175 m * 160 ] / (4.1 x 10⁻³ )

β max = 1.432 x 10 ⁻¹² T

r = 70 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.070 m * 160 ] / (4.1 x 10⁻³ )

β max = 5.726 x 10 ⁻¹² T

r = 105.0 mm

β max = [ π * 60 Hz * 4π x 10⁻⁷ * 8.85 x 10⁻¹² * 0.105 m * 160 ] / (4.1 x 10⁻³ )

β max = 8.589 x 10 ⁻¹² T