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auk411
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Here is the question:
A very thin wire which follows a semicircular curve C of radius R,lies in the upper half of the x-y plane with its center atthe origin. There is a constant current I flowing counter clockwise, starting upward from the end of the wire on the positive x-axis and ending downward at the end on thenegative x axis. The wire is in a uniform magneticfield, which has magnitude B0 and direction parallel to the z-axis in the positive z direction. Determine a symbolic answer in unit-vector notation for the total force on the wire due to the magneticfield. Ignore the forces on the leads that carry the current into the wire at the right end and out of the wire at the leftend.
(Solution check: The numerical value with I =2.00 A, B0 = 3.00 T, and R =4.00 m is 48.0 N in j direction.)
I am not getting the right answer.
First, we know that dF[itex]^{\rightarrow}[/itex] = i dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].
So [itex]\int[/itex]dF[itex]^{\rightarrow}[/itex] = i [itex]\int[/itex]dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].
Then, since B is always perpendicular to L, we have a sin of 90 degrees. This implies that (along with B being constant)
F = iB[itex]\int dL[/itex].
The arclength of a circle is 2[itex]\pi[/itex]R. A semicircle's arclength is [itex]\pi[/itex]R.
Therefore F = iB[itex]\pi[/itex]R.
This doesn't get you to the right answer. What am I doing wrong? Why isn't my answer right? And is there some physical quantity that my answer does correspond to?
A very thin wire which follows a semicircular curve C of radius R,lies in the upper half of the x-y plane with its center atthe origin. There is a constant current I flowing counter clockwise, starting upward from the end of the wire on the positive x-axis and ending downward at the end on thenegative x axis. The wire is in a uniform magneticfield, which has magnitude B0 and direction parallel to the z-axis in the positive z direction. Determine a symbolic answer in unit-vector notation for the total force on the wire due to the magneticfield. Ignore the forces on the leads that carry the current into the wire at the right end and out of the wire at the leftend.
(Solution check: The numerical value with I =2.00 A, B0 = 3.00 T, and R =4.00 m is 48.0 N in j direction.)
I am not getting the right answer.
First, we know that dF[itex]^{\rightarrow}[/itex] = i dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].
So [itex]\int[/itex]dF[itex]^{\rightarrow}[/itex] = i [itex]\int[/itex]dL[itex]^{\rightarrow}[/itex] x B[itex]^{\rightarrow}[/itex].
Then, since B is always perpendicular to L, we have a sin of 90 degrees. This implies that (along with B being constant)
F = iB[itex]\int dL[/itex].
The arclength of a circle is 2[itex]\pi[/itex]R. A semicircle's arclength is [itex]\pi[/itex]R.
Therefore F = iB[itex]\pi[/itex]R.
This doesn't get you to the right answer. What am I doing wrong? Why isn't my answer right? And is there some physical quantity that my answer does correspond to?
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