Physics 202
Tuesday, April 6, 1999
Announcements:
Lecture notes:
Magnetic Field Due to a Solenoid
•
A solenoid is a tightly wound helical coil of wire. See figures 3017 and 3018 on
page 738 of the text for a picture.
•
The magnetic field outside of a solenoid is very, very small.
Refer to figure 3020 from page 739 of the text for the following example
•
To determine the magnitude of B inside the solenoid, draw an amperian loop as shown.
o
I(integral)
B (dot product) ds =
I (integral from a to b)
B (dot product) ds +
I
(integral from b to c)
B (dot product) ds +
I (integral from c to d)
B (dot
product) ds +
I (integral from d to a)
B (dot product) ds
o
B = 0 outside of a solenoid, so
I (integral from c to d)
B (dot product) ds = 0
o
I (integral from b to c)
B (dot product) ds =
I (integral from d to a)
B (dot
product) ds = 0 because the field is perpendicular to ds.
o
B is only found inside the solenoid, so
I(integral)
B (dot product) ds =
I
(integral from a to b)
B (dot product) ds =
μ
o
N i
o
But,
I (integral from a to b)
B (dot product) ds = Bh
o
Therefore, Bh =
μ
o
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 Spring '07
 MAHLON,GREGORYDA
 Physics, Magnetism, Magnetic Field, Faraday, loop

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