Rate of Magnetic Field Change

1. Mar 30, 2006

wr1015

A conducting loop of wire has an area of 6.9 10-4 m2 and a resistance of 110$$\Omega$$ . Perpendicular to the plane of the loop is a magnetic field of strength 0.18 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.18 A?

here's what i've done so far:

$$\theta$$ = 0

i used Ohm's Law V = IR to find V which is the same as EMF, which came out to be 19.8 V. Then to find initial flux i used $$\phi_{i}$$ = (.018T) (6.9 x 10$$^-4$$) (cos 0) and got 1.242 x 10$$^-4$$

this is where i think i might have gone wrong:

i'm assuming they are talking about a change over 1 second, so $$\Delta$$t = 1s

and N = 1 since it originally says "A conducting loop of wire"

so i set up the Emf formula like this: 19.8 V = ($$\phi_{f}$$ - 1.242 x 10$$^-4$$) and solved for $$\phi_{f}$$ and of course i'm not getting the right answer... any suggestions??

Last edited: Mar 30, 2006
2. Mar 30, 2006

Staff: Mentor

Start by writing Faraday's law, which relates induced EMF with the rate of change of the flux through the loop.

3. Mar 30, 2006

wr1015

i did: 19.8 = 1 (($$\phi_{f}$$ - 1.242E-4)/(1))

$$\mid$$E$$\mid$$ = 19.8 V
N = 1
$$\phi_{i}$$ = 1.242E-4 T
$$\Delta$$t = 1s

Last edited: Mar 30, 2006
4. Mar 30, 2006

wr1015

anyone else?

5. Mar 30, 2006

nrqed

You are fiding the final flux but that`s not what they are asking. They want the rate of change of the magnetic field!
Use that ${d \phi \over dt} = A cos \theta {d B \over dt}$. In your case cos theta = 1. Set this equal to the emf induced and dolve for dB/dt and see if you get the correct answer. As far as I can tell, the initial B field is not needed.

Patrick

6. Mar 30, 2006

wr1015

ahh yes I don't why I thought finding the final flux was finding the rate of change thank you for clearing that up