# Faraday's law of electomagnetic induction

A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.60 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 x 10-2 ohm/m. What is the average electrical energy dissipated in the resistance of the wire.

ok so this is what i have:
r= 13cm= .13m
change in t= .45s
B= .60T
restitance per unit length= 3.3 x 10^-2
N= 1

then i tried:
magnetic flux= BAcos= (.60T)(530.9m^2)= 318.6
emf= -N(change in magnetic flux/ change in time)= -1(318.6/.45s)= -708

here is where i get really confused:
R= restitance per unit length (L/A)

does this mean i should do:
R= 3.3 x 10-2 ohm/m (2 pi .13)/(.13^2 pi)???

or is it just R= 3.3 x 10-2 ohm/m (.13m)??? because the units work out this way...

once i figure that out i would do:
I=emf/R ---->
P=I(emf) ------->
E=Pt and that would give me energy

mukundpa
Homework Helper
you are guven resistance per unit length
hence 2Pi r length will have resistance 3.3x10^-2*(2Pi -0.13)

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won't that give me units of ohm/m^2 ? i thought i was supposed to be getting rid of the m's because the R is in units of ohm...

OlderDan
Homework Helper
$$R= 3.3 \times 10^{-2} \left(\frac{ohm}{m}\right) \cdot \left(2 \pi \times 0.13m \right)$$

This has units of ohms. The last term is the circumference of the circle, i.e., the length of the wire.

ok i think i finally understand how to find R now...
R= .027

once i figured that out, i did:
I=emf/R = (-708)/.027 = -26266.1
P=I(emf) = (-26266.1)(-708) = 18596420.37W
E=Pt = (18596420.37W)(.45s) = 8368389.167J and that should give me the electrical energy dissipated in the wire right???

but this problem still isn't working out and i'm not sure where i went wrong

Check your formula for the magnetic flux. Is the magnetic field always at 0.6 Tesla? Also, pay attention to the fact that the question is asking for the average energy dissipated.