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.

 

[kerimek]

I can provide a response to the above content by explaining the concepts of Faraday's Law and a solenoid and how they relate to the given problem.

Faraday's Law is a fundamental principle in electromagnetism that states that a changing magnetic flux through a surface will induce an electric field along the boundary of that surface. In this case, the changing magnetic flux is caused by the oscillating current in the solenoid, and the surface in question is a circular surface with a radius of 1 cm centered on the axis of the solenoid.

A solenoid is a long coil of wire with multiple turns, and it is commonly used to create a uniform magnetic field inside the coil. In this problem, the solenoid has 1000 turns per meter and a radius of 2.00 cm.

To solve this problem, we can use the equations provided in the prompt. Eq. 1 represents Faraday's Law, where \epsilon is the induced electric field, \phi is the magnetic flux, and t is time. Eq. 2 relates the magnetic flux to the magnetic field B and the area A. Eq. 3 gives the relationship between the magnetic field B and the current I in the solenoid, where \mu_0 is the permeability of free space and n is the number of turns per unit length. Finally, Eq. 4 gives the magnitude of the induced electric field at a distance r from the axis of the solenoid.

To find the electric field at r = 1.00 cm, we can plug in the values given in the problem into Eq. 4:

E = \frac{(2.00cm)^2 \mu_0 (1000 turns/m)}{2 (1.00cm)} 500 \pi sin (100\pit)

Simplifying this expression, we get:

E = 2000 \mu_0 \pi sin (100\pit)

Now, we can determine the direction of the electric field by using the right-hand rule. Point your thumb in the direction of the current, which is counterclockwise in this case. Your fingers will curl in the direction of the magnetic field, which is into the page. By applying Faraday's Law, we know that the induced electric field will be perpendicular to both the current and the magnetic field, so it will be pointing out of the page. Therefore, the direction of