Inducing an EMF in an Aluminum Ring

[Color_of_Cyan]

Homework Statement

An aluminum ring of radius r1 = 5.00 cm and a resistance of 2.65 x 10^-4 Ω is placed around one end of a long air-core solenoid with 970 turns per meter and radius r2 = 3.00 cm as shown in the figure. Assume the axial component of the field produced by the solenoid is one-half as strong over the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s.

Homework Equations

ε = d(magnetic flux) / dtmagnetic flux = ∫(B)(dA)I = ε / Rmagnetic field in a solenoid:

B = μ₀(turns / length)(current)Ampere's law? :

∫B ds = μ₀I(+ some other formula / combination I'm probably missing)

The Attempt at a Solution

I think I mostly need help just finding the induced emf (ε). I know you're to somehow find the magnetic flux of what I believe is the ring. Afterwards you find the magnetic flux and you can divide it by time (rather the time interval in which the magnetic flux changes) giving you the induced emf?

But it's complicated because there's changing current instead, which is in the solenoid, and I am not sure what to do about it, or how it relates to the change in flux over time. At the end you divide the induced emf by the resistance of the ring giving you the induced current.

 

[tiny-tim]

Hi Color_of_Cyan!

magnetic field in a solenoid:

B = μ₀(turns / length)(current)
…
I think I mostly need help just finding the induced emf (ε).

I don't see the difficulty …

B = μ₀(turns / length)(current),

so dB/dt = μ₀(turns / length)d(current)/dt​

 

[Color_of_Cyan]

Hi Color_of_Cyan! I don't see the difficulty …

B = μ₀(turns / length)(current),

so dB/dt = μ₀(turns / length)d(current)/dt​

Sorry if I sound dumb:If I solve for flux how do I account for the magnetic field that you now show is changing? (I'm really not too good with calculus either just so you know)I'm thinking since flux = ∫B dA

Then flux = ∫[ (dB/dt)(dA) ] ? (Not sure what to do about this)

On the subject of flux of the RING, the area is still πr² right, even though it's a ring?

I guess the flux is changing as well since the magnetic field is changing. I'm thinking something along the lines that

flux = μ₀(970 turns / meter)[ π(0.05m)²]

then emf is just = μ₀(970 turns / meter)[ π(0.05m)²](270A / time change in s)

?

 

[tiny-tim]

flux = ∫B dA

so d(flux)/dt = ∫[ (dB/dt)(dA) ]

On the subject of flux of the RING, the area is still πr² right, even though it's a ring?

what matters is the flux through the ring

but the question tells you not to bother to calculate it properly …

it tells you to assume that the flux through the ring (radius r₁) is simply the flux through radius r₂, and also that the B field there is half the B field at the centre of the solenoid

 

[Color_of_Cyan]

Yes I meant flux through the ring, my bad.

it tells you to assume that the flux through the ring (radius r₁) is simply the flux through radius r₂, and also that the B field there is half the B field at the centre of the solenoid

I can't see where you got that the flux through the ring is the same as the flux through r₂ though


So, the flux would be the same using the area of the solenoid end though? Also multiply by 1/2 ?

So would it be

flux = ∫B dA

emf = d(flux)/dt

emf = (1/2)(μ₀)(970 turns / meter)(270 A / s)[∏(0.03m)²]

?

 

[tiny-tim]

I can't see where you got that the flux through the ring is the same as the flux through r₂ though

from …

… Assume the axial component of the field produced by the solenoid is one-half as strong over the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area.

since the ring is perpendicular to the axis, and since the field is obviously cylindrically symmetrical, that means only the axial component is relevant to the flux through the ring

and it specifically tells you to ignore any field outside r₂ …

so the flux through the ring (r₁) is the same as the flux through r₂

So, the flux would be the same using the area of the solenoid end though? Also multiply by 1/2 ?

So would it be

flux = ∫B dA

emf = d(flux)/dt

emf = (1/2)(μ₀)(970 turns / meter)(270 A / s)[∏(0.03m)²]

?

looks good!

 

[Color_of_Cyan]

Thanks.