Magnetic Field Along The Axis of A Solenoid
Magnetic Field Along The Axis of A Solenoid
Magnetic Field Along The Axis of A Solenoid
Axis of a Solenoid
AP Physics C
Montwood High School
R. Casao
Consider a solenoid
of length L, radius
R, turns N, carrying
current I.
We will determine
an equation for the
magnetic field B at
an axial point P
inside the solenoid.
Consider the solenoid as a distribution of
current loops.
The magnetic field for any one loop is:
o I R 2
Bx
2 x R2 2
3
2
o R di
2
dB
2x R
3
2 2 2
o R 2
N
dB I dx
2 x R 2 2
3
2 L
For each element of length dx along the length of
the solenoid, the distance x and the angle
change.
The value of R remains constant.
Express x in terms of the angle and find dx.
x
tan x R tan
R
dx d(R tan ) R d(tan )
dx R sec d
2
Substitute:
o R 2 N
dB I R sec2 d
2 R tan R
2 2
3
2 L
o R I N sec d
3 2
dB
2 R tan R
2 2 2
3
2
L
o R I N sec d
3 2
dB
2 R (tan 1)
2 2
3
2
L
o R I N sec d
3 2
dB
2 R tan
2
3
2 2
1 2 L
3
tan 1 sec
2 2
o R I N sec d
3 2
dB
2 R sec
2 3 2
3
2
L
o R I N sec d
3 2
dB
2 R 6
sec
2 3
L
o R I N sec d
3 2
dB
2 R sec L
3 6
o R I N sec d
3 2
dB
2 R sec L
3 6
o I N sec d
2
dB
2 sec L
3
o I N d
dB
2 sec L
o I N cos d
dB
2L
Integrate from 1 to 2:
2 o I N cos d
2
1 dB 1 2L
o I N 2
B cos d
2L 1
o I N
sin
2
B
2L 1
o I N
B sin 2 sin 1
2L
If point P is at the midpoint of the solenoid and if
the solenoid is long in comparison to the radius
R, then 1 = -90 and 2 = 90. The result is the
equation for the magnetic field at the center of a
solenoid.
o I N
B sin 90 sin 90
2L
o I N o I N
B 1 1 2
2L 2L
o I N
B
L
If point P is a point at the end of a long solenoid
towards the bottom, then 1 = 0 and 2 = 90.
The answer shows that the magnetic field at the
end of a solenoid approaches the value at the
center of the solenoid.
o I N
B sin 90 sin 0
2L
o I N o I N
B 1 0 1
2L 2L
o I N
B
2L
Graph of magnetic field B
at axial points vs. distance
x for a solenoid.