Relating Voltage to Magnetic Field

[llenkic]

Homework Statement

I did a lab experiment with Helmholtz coil, where we had to build a probe and connect it to an oscilloscope to measure the induced voltage, from the magnetic field produced by the Helmholtz coil. The Helmholtz coils were connected to an AC power source.

Now I have to relate the induced voltage, to the magnetic field strength inside the coils.

Homework Equations

$$\epsilon$$ = -$$\frac{d\Phi_{B}}{dt}$$

The Attempt at a Solution

I know that by Faraday's Law, that a changing magnetic flux will induce an emf. And since the coils were connected to an alternating current, the current is probably varying sinusoidally.

If this were a situation where the area that was changing with respect to time, I would do this

$$\epsilon$$ = -B$$\frac{dA}{dt}$$

since the magnetic flux, $$\Phi$$_b = BA cos$$\varphi$$.

I feel I should do something like that, but have current varying with time, except I am really having trouble relating the magnetic flux to current...

 

[da_nang]

If I understand correctly, you're inserting a loop of wire as a probe into the coil, then measure the induced voltage to get the magnetic field strength of the coil at that point, yes?

If the loop is inbetween the two coils and parallell to the coil's plane, then the induced voltage follows ε = -dΦ/dt = -BdA/dt. As the magnetic field probably isn't homogenous when you insert it directly, have the loop in the middle of the plane of the coil then rotate it to get a better value for B.

Then you can measure the angle between the normal to the plane of the coil and the normal to the plane of the loop as θ. Then the equation becomes ε = -BA*dcos(θ)/dt.

Now assuming that the loop rotates at a constant speed ω, then ε = -BA*dcos(ωt)/dt.

ε = -BA*dcos(ωt)/dt = BAωsin(ωt) from which B = ε/(Aωsin(ωt)).

Although, if using AC current for the coil, then it would be best not to rotate it. Since the magnetic field created by a Helmholtz coil is B = (4/5)^3/2*μ₀nI/R for DC, AC will have I as i*sin(2πft - φ), where i is the peak current.
Then B = (4/5)^3/2*μ₀ni*sin(2πft - φ)/R = B₀sin(2πft - φ), where B₀ is the peak magnetic field strength.

Thus ε = -dΦ/dt = -AB₀dsin(2πft-φ)/dt = -2πf*AB₀cos(2πft-φ).

 

[llenkic]

Yes, that's what I had to do, the probe was inserted at the axis of the coil.

So you just replaced the current in the DC expression for magnetic field and replaced it with an AC expression for current? I hadn't thought of replacing B in the magnetic flux equation with an expression for the magnetic field... That makes so much sense, thank you for your help!