How Does Faraday's Law Apply to a Solenoid with an Oscillating Current?

[llauren84]

Homework Statement

A long solenoid with 1000 turns per meter and a radius 2.00 cm carries an oscillating current given by I = (5.00A) sin (100 pi t). What is the electric field induced at a radius r = 1.00 cm.from the axis of the solenoid? What is the direction of the electric field when the current is increasing counterclockwise in the coil?

Homework Equations

Eq. 1: $$\epsilon=\frac{-d\phi}{dt}$$

Eq. 2: $$\phi=BA$$Eq. 3: $$B=\mu_0 n \frac{-dI}{dt}$$

Eq. 4: $$E=\frac{\epsilon}{2 \pi r}$$

The Attempt at a Solution

I think that you just sub Eq.3 into Eq.2 and then Eq.2 into Eq.3 and then into Eq 4 and take the derivative of I:

$$E = \frac{R^{2} \mu_0 n}{2 r} 500 \pi sin (100\pit)$$

My confusion is where to put the r=1cm and R=2cm.

Also, how do you know which direction the E field is going?Thanks for your help. =)

 

[Redbelly98]

Homework Equations

Eq. 1: $$\epsilon=\frac{-d\phi}{dt}$$

Agreed.

Eq. 2: $$\phi=BA$$

Agreed. Question for you: what is A here? More specifically, what size loop are you using to get Φ?

Eq. 3: $$B=\mu_0 n \frac{-dI}{dt}$$

Eq. 3 is wrong. Look it up again, what is the B-field inside a solenoid?

Eq. 4: $$E=\frac{\epsilon}{2 \pi r}$$

Agreed. Question for you: what is r here? More specifically, what size circle are you using to relate ε and E?

The Attempt at a Solution

I think that you just sub Eq.3 into Eq.2 and then Eq.2 into Eq.3 and then into Eq 4 and take the derivative of I:

$$E = \frac{R^{2} \mu_0 n}{2 r} 500 \pi sin (100\pit)$$

What happened to Eq. 1? I think you better show what steps you took to get this equation.

My confusion is where to put the r=1cm and R=2cm.

See my comments after Eq's 2 and 4.

Also, how do you know which direction the E field is going?

Lenz's Law is useful here.