cont'd... when perpendicular to a 0.81 T magnetic field?
N/m
b.) What if the angle between the wire and field is 60.0 degrees?
N/m
I would appreciate ANY help w/ this problem! Thanks!
2007-02-19 09:58:32 · 4 answers · asked by Jade m 1 in Science & Mathematics ➔ Physics
(a)9.79x0.81 N/m (b)9.79x0.81xsin 60 degree N/m
2007-02-27 02:44:34 · answer #1 · answered by ukmudgal 6 · 2⤊ 1⤋
From Ampere's regulation, the magnetic container precipitated by ability of the 12 A present day is given by ability of 2pi r B =uo Ienc B = uo Ienc / (2pi r) = in the path perpendicular to the twine.The expression for the lorentz stress that ends up in the different twine is F = q v X B or F/length = I B the place F/length is the stress consistent with unit length. So your answer is I = F/length / B = F/length / (uo Ienc / (2pi r)) the place F/length is given, uo is the permeability in vacuum, Ienc = 12 A, and r = 6cm. Watch with the gadgets. Checking on the vectors and the bypass product, the present has to flow vertically, only like quite a few different present day.
2016-11-23 19:24:21 · answer #2 · answered by Anonymous · 0⤊ 0⤋
This may be a little odd but I think it is or has to do with the fact on the current of amps in mini amps or standard amps. I might have to do with how thick the wire is and whether the 9.79 is anywhere to 100 volts. this part I'm not sure.
2007-02-19 10:08:53 · answer #3 · answered by GM Waldron 1 · 0⤊ 0⤋
the net gravity pull is still 9.3 m/s/s--the towers can withold twenty degree bend at 275, 000 volts. At high tension they can transmit 1 million volts at a resistor strength of kick back transformers.
The electromagnetic field is small.
2007-02-19 10:16:26 · answer #4 · answered by Kilty 5 · 0⤊ 0⤋