Calculating Induced EMF in a Larger Solenoid Slipped Over Another Solenoid

[cuallito]

1. Problem statement

A very long straight solenoid has a diameter of 3cm, 40 turns per cm, and a current of .275 A. A second solenoid is with larger diameter is slipped over it with N turns per cm, and the current is ramped down to zero over 0.2 s.

a) What is the emf induced in the second coil if it has a diameter of 3.6cm and N=6?
b) The diameter is 7.2cm and N=12 ?

Homework Equations

B=μNI

The Attempt at a Solution

Wouldn't the answer to both (a) and (b) be zero, since the magnetic field outside a solenoid is basically zero?

 

[TSny]

Wouldn't the answer to both (a) and (b) be zero, since the magnetic field outside a solenoid is basically zero?

No. An emf will be induced in the second solenoid as long as there is a changing magnetic flux through it. Pretty amazing. The changing magnetic field of the first solenoid induces an electric field both inside and outside the first solenoid. This induced electric field is what creates the emf in the second solenoid. But you don't need to calculate this electric field in order to work the problem.

Is the diameter of the first solenoid given?

 

[cuallito]

Oh sorry, left it it out, 3cm.

So would the magnetic flux passing thru both solenoids be the same, like a transformer?

So let me take a stab at it here:

The magnetic flux from the first solenoid would be:

Φ1=BA
A=cross sectional area
B=μ*N1*I
N1=number of turns of 1st solenoid
Φ1=μ*N1*I*A

Magnetic flux thru both solenoids would be the same Φ1=Φ2

Ampere's law for 2nd solenoid:
emf=-N2*Φ/Δt
emf=-N2*μ*N1*I*A/Δt

Plugging in for (a) I get -2.93*10^-7 V
for (b) I get -5.86255*10^-7

Is that right?

 

[TSny]

Φ1=BA
A=cross sectional area
B=μ*N1*I
N1=number of turns of 1st solenoid
Φ1=μ*N1*I*A

Note that N1 is the number of turns per unit length of solenoid 1

Magnetic flux thru both solenoids would be the same Φ1=Φ2

Ampere's law for 2nd solenoid:
emf=-N2*Φ/Δt
emf=-N2*μ*N1*I*A/Δt

OK. (Faraday's law - not Ampere's law)

Plugging in for (a) I get -2.93*10^-7 V
for (b) I get -5.86255*10^-7

I do not get the same power of 10. Did you express N1 in SI units?

 

[cuallito]

Ah, so N1 would be 40/.01 = 4000 turns per meter.

So you'd get to the -5 instead of -7th power?

 

[TSny]

Ah, so N1 would be 40/.01 = 4000 turns per meter.

So you'd get to the -5 instead of -7th power?

Yes, that is what I got.