# Ampere's circuital law

1. Aug 8, 2012

### zhillyz

1. The problem statement, all variables and given/known data

State Ampere's circuital law. The figure shows a region A in which the magnetic field $H$ and of value 20Am^-1 to the right and a region B in which the magnetic field $H$ 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.

2. Relevant equations

$\int_c H\cdot dl$$= I_{enc}$

3. 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..

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

Or am I missing something?

2. Aug 8, 2012

### tiny-tim

hi zhillyz!

(try using the X2 button just above the Reply box )
i don't understand

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?

3. Aug 8, 2012

### zhillyz

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.

4. Aug 8, 2012

### tiny-tim

ah!
what does the "dot" do to the "vertical" bits?

(and you'll need an H1 and an H2)

5. Aug 9, 2012

### zhillyz

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..

$I_{1enc} = H_1 \cdot y = 20\cdot 0.05 = 1A$
$I_{2enc} = H_2 \cdot y = 30\cdot 0.05 = 1.5A$

Seem correct?

6. Aug 9, 2012

### tiny-tim

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

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

otherwise correct

7. Aug 9, 2012

### zhillyz

Thank you very kindly :).