Magnetic Field Line Integral Problem

AI Thread Summary
The discussion revolves around calculating the line integral of the magnetic field B between two points using Ampere's Law. It is established that the contribution of B from the straight segments is zero due to the perpendicular orientation, while the semicircular segment contributes significantly since its dl component is parallel to B. The integral is expressed as B multiplied by πr, with r given as 0.01 meters, and the right side of the equation is noted as 2A multiplied by μ0. There is confusion regarding the enclosed current, with a need to clarify whether it is 2A or 1A, emphasizing that the Amperian loop must be closed. The conclusion is that the integral on the semicircle is half of the full circle due to symmetry.
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Homework Statement



What is the line integral of B between points i and f in the figure?

knight_Figure_32_22.jpg


Homework Equations



Ampere's Law: ∫B∙dl = Ienclosed * μ0
note: the integral on the left is a line integral.

The Attempt at a Solution



I applied Ampere's Law. I know that the contribution of B to the line integral on the straight segments to the left and right of the semicircle is zero since the magnetic field is perpendicular to the surface (line) at these segments. The semicircle, however, has a dl component that is parallel to the B field at all points along the semicircle. Therefore the integral should be B*pi*r (where r = 0.01 meters) and the right side of the equation should be 2A * μ0. I am confused as to whether the problem is asking for the value of the magnetic field, or whether I am supposed to simply give the value of ∫B∙dl.
 
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They want the integral.
 
Remaining confusion about problem

pam said:
They want the integral.

In that case, the integral is equal to the current enclosed by the surface, multiplied by the constant (mu). Is the value of the current enclosed 2A or 1A? This is a bit unclear to me due to the nature of the Amperian surface. Thank you for helping.
 
The Amperian loop must be closed. Take it as a full circle for Ampere's law. Then, because of the symmetry, the integral on the half circle in the picture is half the full integral.
 
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