Asked by Anonymous
A thin 2.84 m long copper rod in a uniform magnetic field has a mass of 53.7 g. When the rod carries a current of 0.273 A directed perpendicular to the magnetic field, it floats in the magnetic field. The acceleration of gravity is 9.81 m/s2 . What is the field strength of the magnetic field? Answer in units of T.
I thought I had to find force first, so I did using F=ma.
F=ma
F=53.7(9.81)
F=526.797
That didn't seem right because it was such a large number, but I continued with the problem anyway. Using the formula F=ILB (force=current*length*magnetic field vector):
F=ILB
526.797=0.273(2.84)T
And then I solved it algebraically and got 679.457 T, which as expected, wasn't the right answer.
Can anyone help?
I thought I had to find force first, so I did using F=ma.
F=ma
F=53.7(9.81)
F=526.797
That didn't seem right because it was such a large number, but I continued with the problem anyway. Using the formula F=ILB (force=current*length*magnetic field vector):
F=ILB
526.797=0.273(2.84)T
And then I solved it algebraically and got 679.457 T, which as expected, wasn't the right answer.
Can anyone help?
Answers
Answered by
bobpursley
The mass is .0537kg
F=ma=9.8*.0537=5.27N=
ILB=0.273(2.84)T
T=5.27/.775=6.8tesla
F=ma=9.8*.0537=5.27N=
ILB=0.273(2.84)T
T=5.27/.775=6.8tesla
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