The solenoid

The magnetic field along the axis of a solenoid

Objective:

Students will use a spreadsheet to calculate the magnetic field along the x-axis of a solenoid of length ls, radius R, and n turns per unit length carrying current I.  The solenoid is centered at the origin with its axis along the x-axis.  The field at any point x is

B = (1/2)m0nI[f1(x)-f2(x)],

where

f1(x) = (x+(1/2)ls) / [R2+(x+(1/2)ls)2]1/2

and

f2(x) = ( x-(1/2)ls) / [R2+(x-(1/2)ls)2]1/2.

Procedure:

(a)  Let ls=10cm, R = 0.5cm I = 4A, and n = 1500 turns/m.  Construct a spreadsheet that calculates B as a function of x for positive values of x ranging from x = 0 to x = 30cm in steps of 0.2cm.  Plot B versus x.  (Note:  use SI units consistently.)

(b)  Change R to the following values and observe the effects on the graph in each case:  R = 0.1cm, 1cm, 2cm, 5cm, and 10cm.  For small values of R (small compared to ls) the field inside the solenoid should look like that of an ideal (infinitely long) solenoid.  For large values of R the field should look like that of a loop.  Does it?

(c)  For x >> ls (where x is measured from the center of the solenoid)  the magnetic field of a finite solenoid along the x-axis approaches

B(x) = (m0 / 4p)(2m / x3)

where m = NIpR2 is the magnetic dipole moment of the solenoid.  Modify your spreadsheet  to calculate and plot x3B(x) as a function of x.  [Multiply your column containing B by the cube of your column containing x.}  Verify that x3B(x) approaches a constant for x >> ls.  [Use a log scale for the y-axis.   Leave out the first point since the log of zero is not defined.]

To earn extra credit, save your Excel document (your name_exm7.xls) and attach it to an e-mail message to mbreinig@utk.edu.  Comment on your results.