Magnitude of the non-coulomb electric field around a solenoid.

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SUMMARY

The discussion centers on calculating the magnitude of the non-Coulomb electric field around a solenoid with a metal ring. The solenoid has 10,000 turns and experiences a magnetic field change from 5 T to 3 T over 0.1 seconds. The initial calculation for the electromotive force (emf) yielded 251.3 volts, leading to an electric field of 166.7 V/m. However, the correct approach involves recalculating the emf from the perspective of the metal ring, resulting in an emf of 3.62 volts and an electric field of 2.40 V/m.

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  • Basic concepts of solenoids and magnetic fields
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Pepjag
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Homework Statement



A metal ring of radius r = 24 cm surrounds a
solenoid of radius R = 2 cm and 10000 turns.
The magnetic field inside the solenoid changes from 5 T to 3 T in
0.1 s. Find the magnitude of the non-Coulomb
electric field along the ring.
Answer in units of V/m

Homework Equations



emf = -N*d(flux)/dt = -N*db/dt*A
emf = integral of E*dl

The Attempt at a Solution



Ok, so i pretty much just plugged in numbers. for the emf:

emf = -10000*(3-5)/0.1*pi*0.02^2 = 251.3 volts.

For the electric field:

E = emf/L = 251.3/2*pi*0.24 = 166.7 V/m

Unfortunately, the computer keeps saying I'm wrong. Am I really doing something wrong? This seemed pretty straight forward.
 
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Pepjag said:
Ok, so i pretty much just plugged in numbers. for the emf:

emf = -10000*(3-5)/0.1*pi*0.02^2 = 251.3 volts.

Am I really doing something wrong?

You're looking at the emf induced in the metal ring. How many turns of wire does a metal ring have?
 
Wait, so you have to use the emf equation from the ring's perspective?

emf = (3-5)/0.1*pi*0.24^2 = 3.62 volts.

E = emf/L = 251.3/2*pi*0.24 = 2.40 V/m

Or do I still use the solenoid's area for the emf?
 
Last edited:
I have until 11:30 ct to finish this one problem, can anyone please give input.
 

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