- #1
physics604
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1. A coil of 25 turns of wire is suspended by a thread. When a current flows through the coil, the tension in the thread is reduced by 4.0 x 10-2 N. What are the magnitude and direction of the current? {magnitude of current; direction of current}
a) 4.1 A; clockwise
b) 0.16 A; clockwise
c) 0.16 A; counter-clockwise
d) 4.1 A; counter-clockwise
F=IlB
B= [itex]\frac{\mu NI}{l}[/itex]
B=[itex]\mu[/itex]NI/l
F= IlB = Il([itex]\mu[/itex]NI/l) = I2[itex]\mu[/itex]N
I = [itex]\sqrt{ \frac{F}{\mu N} }[/itex]
so I = √4.0x10-2 / 4∏x10-7 x25 = 35.48A
My answer doesn't match with any of theirs.
Also, how would I find the direction of current? I don't know the direction of B (magnetic field).
Any help would be greatly appreciated.
a) 4.1 A; clockwise
b) 0.16 A; clockwise
c) 0.16 A; counter-clockwise
d) 4.1 A; counter-clockwise
Homework Equations
F=IlB
B= [itex]\frac{\mu NI}{l}[/itex]
The Attempt at a Solution
B=[itex]\mu[/itex]NI/l
F= IlB = Il([itex]\mu[/itex]NI/l) = I2[itex]\mu[/itex]N
I = [itex]\sqrt{ \frac{F}{\mu N} }[/itex]
so I = √4.0x10-2 / 4∏x10-7 x25 = 35.48A
My answer doesn't match with any of theirs.
Also, how would I find the direction of current? I don't know the direction of B (magnetic field).
Any help would be greatly appreciated.