- #1
mayo2kett
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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
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