Induced Electric Fields of metal ring

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aznkid310
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Homework Statement



A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s.

(a) What is the magnitude of the electric field induced in the ring?

(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

Homework Equations



Do i need to do anything with the initial B value?

Change in flux dB/dt = -0.25t?

Or is it dB/dt = 1.12 - 0.25t?

The Attempt at a Solution



a) d[phi]/dt = (dB/dt)*Acos(0) = (-0.250)*(pi(2.25*10^-2)^2) = -3.98*10^-4 Wb

E = (1/2r*pi)*(d[phi]/dt) = -2.8*10^-3 N/C

b) Clockwise?
 
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induced emf = - (change in magnetic flux)

Since magnetic flux is decreasing, an induced magnetic field opposite to that must be created to counteract this change in flux. Is my reasoning off?

Also, is part (a) correct?
 
Your answer to part a looks right to me (except they want the magnitude of the field, so you don't need the negative sign).

aznkid310 said:
induced emf = - (change in magnetic flux)

Since magnetic flux is decreasing, an induced magnetic field opposite to that must be created to counteract this change in flux. Is my reasoning off?

Also, is part (a) correct?

The induced magnetic field will be in the direction to oppose the change. Since the magnetic flux from the magnets is decreasing, the induced magnetic field will be in the same direction as the magnet's field. Does that make sense?
 
Ah that makes sense. It would oppose only if the magnetic flux is incresing right?
 
It always opposes the change, so if the magnetic flux is increasing, then yes, the induced field will be in the opposite direction as the external flux.