Maximum electric field from magnetic field

[Schfra]

Homework Statement

In the central region of a solenoid that is connected to a radio frequency power source, the magnetic field oscillates at 2.5·10^6 cycles per second with an amplitude of 4 gauss. What is the amplitude of the oscillating electric field at a point 3 cm from the axis? (This point lies within the region where the magnetic field is nearly uniform.)

Homework Equations

∇ × E = −∂B/∂t (maybe)

The Attempt at a Solution

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I could only find one equation relating the electric field to the magnetic field in my book. I’m not sure if it’s right or not.

∇ × E = −∂B/∂t

I’m not really sure where to go from here. I’m guessing I have the wrong equation, but I’m not quite sure since I haven’t taken calc 3 and am not very familiar with partial differentiation or gradients.

 

[vela]

What book are you using? Usually, you start with the integral version of that law. (And it is the one you want to use.) Is that in your book?

 

[Schfra]

What book are you using? Usually, you start with the integral version of that law. (And it is the one you want to use.) Is that in your book?

Is it this one?

I’m not quite sure how to use this equation. It was listed first, I just didn’t include it because I didn’t see the electric field anywhere.

 

[vela]

Yeah, that's the one, though it's not in the most useful form. If you integrate the differential form of the law on a surface $S$ bounded by a closed path $C$, you get
$$\int_S (\nabla \times \vec{E})\cdot d\vec{A} = -\frac{\partial}{\partial t} \int_S \vec{B} \cdot d\vec{A}.$$ The righthand side is just the rate of change of the magnetic flux through the surface $S$. Using Stokes' theorem, you can express the lefthand side as a line integral:
$$\oint_S (\nabla \times \vec{E})\cdot d\vec{A} = \oint_C \vec{E}\cdot d\vec{l}.$$ (This is what you will learn in Calc 3, but you'll just have to accept it right now as a given.) Combining the two, you get
$$\oint_C \vec{E}\cdot d\vec{l} = -\frac{\partial}{\partial t} \int_S \vec{B} \cdot d\vec{A}.$$ This is the form that's more useful for solving this problem. The lefthand side you should recognize as the induced emf around the path $C$. Using the symmetry of the situation in the problem, you can evaluate the line integral for the appropriate path.

 

[Schfra]

Yeah, that's the one, though it's not in the most useful form. If you integrate the differential form of the law on a surface $S$ bounded by a closed path $C$, you get
$$\int_S (\nabla \times \vec{E})\cdot d\vec{A} = -\frac{\partial}{\partial t} \int_S \vec{B} \cdot d\vec{A}.$$ The righthand side is just the rate of change of the magnetic flux through the surface $S$. Using Stokes' theorem, you can express the lefthand side as a line integral:
$$\oint_S (\nabla \times \vec{E})\cdot d\vec{A} = \oint_C \vec{E}\cdot d\vec{l}.$$ (This is what you will learn in Calc 3, but you'll just have to accept it right now as a given.) Combining the two, you get
$$\oint_C \vec{E}\cdot d\vec{l} = -\frac{\partial}{\partial t} \int_S \vec{B} \cdot d\vec{A}.$$ This is the form that's more useful for solving this problem. The lefthand side you should recognize as the induced emf around the path $C$. Using the symmetry of the situation in the problem, you can evaluate the line integral for the appropriate path.

Would the left side of that equation be equal to the electric field multiplied by the circumference of a circle that has a radius of the .03m, the distance from the axis?
E(2πr)
=.06(pi)E

 

[vela]

Yes.

 

[Schfra]

Yes.

For the right side I assume B is .0004T. I’m not quite sure about what surface I’m integrating over or about the math there though. What surface is the integral referring to?

 

[vela]

The surface is the disk defined by the circle.

 

[Schfra]

The surface is the disk defined by the circle.

So would the integral on the right hand side be equal to
π(.03m)^2(.0004T)
= 3.6E-7π

And wouldn’t the partial derivative of that be equal to 0?

 

[vela]

The problem statement says the magnetic field oscillates with time, so you don’t want to set B to a constant.

 

[Schfra]

The problem statement says the magnetic field oscillates with time, so you don’t want to set B to a constant.

How would I find that then? I’m guessing that 2.5E6hz and .0004T will both be part of it.

Would it maybe be .0004sin(2.5E6t)?

 

[vela]

Almost. Remember the argument of the sine function needs to be in radians, not cycles.

 

[Schfra]

Almost. Remember the argument of the sine function needs to be in radians, not cycles.

So would it be .0004sin(2pi(2.5E6t))?

I believe sin(2pi) gives a frequncy of one cycle per second. Then multiplying the inside by 2.5E6 should also multiply the frequency by the same number.