example: calculate the magnetic field at point o due to the wire segment shown. the wire carries...

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Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . I C A R Thanks to Dr. Waddill for the use of the diagram. O Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle . I see three “parts” to the wire. As usual, break the problem up into simpler parts. A’ to A A to C C to C’

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Page 1: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

I

C

A

R

Thanks to Dr. Waddill for the use of the diagram.

O

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

I see three “parts” to the wire.

As usual, break the problem up into simpler parts.

A’ to A

A to C

C to C’

Page 2: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

I

C

A

R

O

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

For segment A’ to A: dsr

ˆ ˆ 0

ds r =ds r sin =0

A'AdB =0

A'AB =0

ˆμ

02

I ds rdB=

r

Page 3: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

C

A

R

O

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

For segment C to C’: dsr

ˆ ˆ

ds r =ds r sin 180 =0

CC'dB =0

CC'B =0

ˆμ

02

I ds rdB=

r

I

Page 4: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two straight segments and a circular arc of radius R that subtends angle .

C

A

R

Important technique, handy for homework and exams:

The magnetic field due to wire segments A’A and CC’ is zero because ds is either parallel or antiparallel to along those paths.

rO

dsr

ds

r

Page 5: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight segments and a circular arc of radius R that subtends angle .

For segment A to C:

ˆ ˆ

ds r =ds r sin 90 =0

ˆμ

02

I ds rdB=

r

C

A

R

O

Ir

ds

=ds 1 1

=ds

μ

0

AC AC 2

I dsdB =dB =

R

Page 6: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

C

A

R

O

Ir

ds

μ

4π0

AC 2

I dsdB =

R

μ

4π 0AC AC 2

arc arc

I dsB = dB =

R

μθ

4π0

AC

IB =

R

μ

4π 0AC 2

arc

IB = ds

R

The integral of ds is just the arc length; just use that if you already know it.

μθ

4π 0AC 2

arc

IB = R d

R

μθ

4π 0AC

arc

IB = d

R

We still need to provide the direction of the magnetic field.

Page 7: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

C

A

R

O

Ir

ds

μ θ ˆ4π

0

A'A AC CC'

IB = B + B + B =- k

R

If we use the standard xyz axes, the direction is ˆ-k.

Cross into . The direction is “into” the page, or .

ds r

x

y

z

Page 8: Example: calculate the magnetic field at point O due to the wire segment shown. The wire carries uniform current I, and consists of two radial straight

C

A

R

O

Ir

ds

Important technique, handy for exams:

Along path AC, ds is perpendicular to .

r

ˆ ˆ

ds r = ds r sin 90

ˆ

ds r =ds