Magnetic Field inside Solenoid and EMF

[jderm]

A small loop of area A is placed inside a long solenoid that has n turns per meter and carries a sinusoidally varying current of amplitude i. The central axes of the loop and solenoid coincide. If i = i₀sin ωt, find the emf in the loop.




EMF_induced = - d($$\phi$$_B)/ dt
$$\phi$$_B=BA
B_solenoid= $$\mu$$₀ni





The product of B and A = A$$\mu$$₀ni₀sin (ωt)
So the derivative of $$\phi$$_B with respect to time:
=A$$\mu$$₀ni₀cos (ωt) ω
so;
EMF_induced = - A$$\mu$$₀ni₀cos (ωt) ω

Apparently this is incorrect. I'm sorry about the sloppy formatting, could someone help me out with this?

 

[Illuminitwit]

EMF = nABω sinωt
F(magnetism) = B • i • l
F(mag) = qvB

I don't know if that helps much, but those are a few more equations that might apply, although I think you already had them in various forms. If you had the magnitude of electrical/magnetic force or the length it might be easier. Sorry...

 

[nlightner]

No problem, let's try to clarify the equations and provide a more accurate response. The magnetic field inside a solenoid is given by B = \mu0ni, where \mu0 is the permeability of free space, n is the number of turns per meter, and i is the current. Since the loop is placed inside the solenoid, the magnetic field passing through the loop is equal to the magnetic field inside the solenoid, Bloop = Bsolendoid = \mu0ni.

The emf induced in the loop is given by Faraday's law, EMFinduced = - d(\phiB)/ dt, where \phiB is the magnetic flux through the loop. The magnetic flux through the loop can be calculated as the product of the magnetic field passing through the loop and the area of the loop, \phiB = BA. Substituting Bloop = \mu0ni and i = i0sin ωt, we get:

EMFinduced = - d(\phiB)/ dt = - d(BloopA)/ dt = - d(\mu0niA)/ dt = - A\mu0n d(i0sin ωt)/ dt

Since the derivative of sin ωt is cos ωt, the final expression for the emf induced in the loop is:

EMFinduced = A\mu0nωi0cos ωt

This means that the emf induced in the loop will also have a sinusoidal variation with the same frequency as the current in the solenoid, but with a different amplitude. It is important to note that this calculation assumes that the loop is small compared to the solenoid and that the magnetic field is uniform within the loop. If these assumptions are not met, the calculation may need to be modified.