Magnetic Field Along the 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:

• The net magnetic field in the solenoid is the sum of the magnetic fields of all the loops.

23

22

2o

x

Rx2

RIμB

• Divide the length of the solenoid into small elements of length dx.– The number of turns in a length dx is:

– The amount of current in an element of length dx is:

– The total current in a length dx is:

dN N NdN = dx

dx L L

dxLN

I

di N I N I= di = dx

dx l l

• The magnetic field contribution dB at point P due to each element dx carrying current di is:

dx

LN

IRx2

RμdB

23

22

2o

2

o3

2 2 2

μ R didB

2 x R

• 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.

2

xtan x R tan

Rdx d(R tan ) R d(tan )

dx R sec d

• Substitute:

dsecR

LN

IRtanR2

RμdB 2

23

22

2o

L1)(tan2

dsecNIRμdB

LRtan2

dsecNIRμdB

23

22

23o

23

222

23o

R

R

LsecR2

dsecNIRμdB

Lsec2

dsecNIRμdB

Lsec2

dsecNIRμdB

sec1tan

L1tan2

dsecNIRμdB

63

23o

326

23o

23

232

23o

22

23

223

2

23o

R

R

R

L2dcosNIμ

dB

Lsec2dNIμ

dB

Lsec2dsecNIμ

dB

LsecR2

dsecNIRμdB

o

o

3

2o

63

23o

• Integrate from 1 to 2:

12o

o

o

o

sinsinL2NIμ

B

sinL2NIμ

B

dcosL2NIμ

B

L2dcosNIμ

dB

2

1

2

1

2

1

2

1

• 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.

LNIμ

B

2L2NIμ

11L2NIμ

B

90sin90sinL2NIμ

B

o

oo

o

• 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 ½ the value at the center of the solenoid.

L2NIμ

B

1L2NIμ

01L2NIμ

B

0sin90sinL2NIμ

B

o

oo

o

• Graph of magnetic field B at axial points vs. distance x for a solenoid.