Posted by **Anonymous** on Saturday, March 15, 2008 at 6:12pm.

An 8m long copper wire with a cross-sectional area of 1.0 x 10^-4 m^2, in the sshape of square loop, is connected to a .10-V battery. If the loop is placed in a uniform magnetic field of magnitude .4T, what is the maximum torque that can act on it? The resistivity of copper is 1.7x10^-8 omega*m (answer: 120 N*m) I can't seem to get the answer. Any help is appreciated!

work:

I=V/R=(.10V)/((1.7x10^-8)/8m))

torque=BIA=(.4T)(I)(10^-4 m^2)

- physics -
**drwls**, Sunday, March 16, 2008 at 6:38am
You have not used the wire resistance R in the Ohm's law equation; you used resistivity, which must be multiplied bu L/A*, where A* is the cross sectional area of the wire (1.0 x 10^-4 m^2) and L is its total length (8 m). The area A that should go into your second equation is the coil area, (L/4)^2 = 4 m^2, since is it s square.

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