What is the current enclosed by a rectangular loop with given magnetic fields?

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Homework Statement



State Ampere's circuital law. The figure shows a region A in which the magnetic field [itex]H[/itex] and of value 20Am^-1 to the right and a region B in which the magnetic field [itex]H[/itex] is uniform and of value 30Am^-1 to the left. Find the current enclosed by the dashed rectangular loop, which has dimensions 8cm x 5cm.

Homework Equations



[itex]\int_c H\cdot dl[/itex][itex]= I_{enc}[/itex]


The Attempt at a Solution



So we already have a constant value for H for both segments and they are constant so I think its adding together two different values. So..

[itex]H \int \cdot dl = I_{enc} = H * (2x + y)[/itex] for x = 4cm and y = 5cm?

Or am I missing something?
 
hi zhillyz! :smile:

(try using the X2 button just above the Reply box :wink:)
zhillyz said:
State Ampere's circuital law. The figure shows a region A in which the magnetic field [itex]H[/itex] and of value 20Am^-1 to the right and a region B in which the magnetic field [itex]H[/itex] is uniform and of value 30Am^-1 to the left. Find the current enclosed by the dashed rectangular loop, which has dimensions 8cm x 5cm.

i don't understand :redface: …

where is the electric wire (for the current)?

what shapes are A and B, and where are they in relation to each other and the wire, and is anything moving? :confused:
 
Apologies perhaps the image itself will clarify and yes I have been that busy brushing up my Latex I forgot the more basic tools were there lol.

Scanned exam question diagram showing a horizontal boundary between region A, with magnetic field H of 20 A/m pointing right, and region B, with H of 30 A/m pointing left, with a dashed rectangular loop straddling the boundary.
 
ah! :smile:
zhillyz said:
So we already have a constant value for H for both segments and they are constant so I think its adding together two different values. So..

[itex]H \int \cdot dl = I_{enc} = H * (2x + y)[/itex] for x = 4cm and y = 5cm?

Or am I missing something?

what does the "dot" do to the "vertical" bits? :wink:

(and you'll need an H1 and an H2)
 
For each half of the rectangles vertical sides the magnetic field moves in equal an opposite directions and therefore would cancel each other out? So it is just..

[itex]I_{1enc} = H_1 \cdot y = 20\cdot 0.05 = 1A[/itex]
[itex]I_{2enc} = H_2 \cdot y = 30\cdot 0.05 = 1.5A[/itex]

Seem correct?
 
hi zhillyz! :smile:
zhillyz said:
For each half of the rectangles vertical sides the magnetic field moves in equal an opposite directions and therefore would cancel each other out? So it is just..

no, you're missing the point (or the dot) …

H.dl is 0 because H is perpendicular to that part of the circuit! :rolleyes:

otherwise correct :smile:​