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 * 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).

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

we need to know the time interval during which the force acts. impulse of force = momentum change

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .