Induced E-field of solenoid, dI/dt and dB/dt not given?

Question

Solenoid

diameter = 7.56 cm, length = 22 cm, N = 189 turns.
Inductance = 0.916 mH

Current at t=infinity is 0.0128 A
Magnetic field at t=infinity is 1.38 x 10^-5 T inside solenoid.

Current as function of time: I(t) = 0.0128A (1 - e^[-t/3.66x10^-6 s])

Magnetic field at time t=1.2 micro seconds: 3.86 x 10^-6 T

So I got this far in the multistep problem, now I have to figure out the induced electric field at a point that is 2 cm radially away from the center of the solenoid of diameter 7.56, and the E field at a point that is 50 cm away from the center. Both at time t=1.2 micro seconds. The answers given are 0.0271 V/m for the 2 cm radius, and 0.00388 V/m at 50 cm radius, but I can't figure out how to get those values. There is an equation given on the worksheet that appears to be a simplification of an equation I have in my book,

Book's equation: E[2(pi)r] = (pi)(r^2)(dB/dt) (in the examples dB/dt is always given but its not here that's my problem...)

Professor's equation: E= {[(Uo)(N/l)(I final)]/[2(pi)(r)tau]}{[e^(-t/tau)][pi][r^2]}
which is read as: permeability constant "Uo" ([4pi]E-7) times number of turns per length times the final current, all divided by 2 pi radius times the time constant "tau". This fraction is multiplied by pi times radius squared times e to the negative of time over "tau". The answer is 0.0271 V/m. Where does he get this equation from?

His equation for the induced electric field outside of the solenoid is the same except instead the last expression is (D/2)^2 instead of r^2. I don't get it...help!

Answer 1

There are different approaches, and you have been given more data than you need. Here's how I would do it.



It will be neater to write:

I(t) = 0.0128A (1 - e^[-t/3.66x10^-6s])

in the form

I(t) = I₀[1 - e^(-t/τ) ]



where

I₀ = 0.0128A

τ = 3.66x10⁻⁶s



Differentiating:

dI/dt = (I₀/τ)e^(-t/τ) (equation 1)



The magnetic field in the coil is proportional to the current: B = kI. We can get, k, the constant of proportionality from the geometrical data given. But the easiest method is to use the fact that B = 1.38 x 10⁻⁵ T when I = 0.0128 A. k = B/I so:

k = (1.38 x 10⁻⁵ T)/(0.0128 A) = 1.078x10⁻³ T/A



B = kI therefore

dB/dt = k dI/dt



From equation 1:

dB/dt = k(I₀/τ)e^(-t/τ) (equation 2)



From equation 2, when t = 1.2μs:

dB/dt = (1.078x10⁻³ x 0.0128 / (3.66x10^-6 ))e^(-1.2X10⁻⁶/(3.66x10⁻⁶))

= 2.716 T/s



Your text book equation:

2πrE = πr²(dB/dt)

is only true for points INSIDE the coil. It is from the integral form of Faraday's Law. It simplifies to:

E = ½r(dB/dt)



When r = 0.02m

= ½ x 0.02 x 2.716

= 0.0272 V/m (minor rounding error)



When r = 0.50m we are outside the coil and the area enclosed by the current-loop is the area of the coil (radius R, so area = πR²) giving:

2πrE = πR²(dB/dt)

E = ½(R²/r)(dB/dt)



R = 7.56/2 cm = 0.0378m

When r = 0.50m

E = ½ x (0.0378² / 0.50) x 2.716

= 0.00388 V/m

Answer 2

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Answer 3

ok r=D/2 if r is radius and D is diameter



Profs equ look like it is just working out the current/time slope at time t given the time constant tau and the final current then getting the field at that time from the magfield/time slope.