Problem 2:

A certain wire carries a current I and is long and straight except for a semicircular loop of radius R. Consider the magnetic field at the center of the semicircle (center being defined as where the center would be if the circle were whole).
  1. Explain why the Biot Savart law tells us that the contribution from the long straight sections is zero.
  2. What is the direction of the magnetic field at the center of the semicircle?
  3. What is the magnitude of the magnetic field at the center of the semicircle, both as a formula and in Tesla if R = 5 cm and I = 2 A?
Solution:

  1. The Biot-Savart law is    dB = (µ0/4) I dl x r/r2

    (here r is a unit vector). Define the x-direction as to the right, y-direction as up, and z-direction as out of the page. For any location on the long straight sections, the vector Idl is in the x-direction, and the unit vector r which points from the source of the field (a piece of the straight section) to the field point (the center of the semicircle) is either in the x-direction or the -x direction. Thus the cross product I dl x r is zero (cross product of two parallel or anti-parallel vectors is zero), and so B from these sections is zero (at the center).

  2. Use the right-hand rule. For example, line your thumb up with the current direction somewhere in the semicircle, and your fingers will curl into the page on the inside of the circle. Thus the field at the center is into the page.

    Or, use the Biot-Savart law. Consider a piece of the semicircle. Now I dl is tangent to the circle, pointing clockwise, and r is pointing radially inward. Thus their cross product points into the page.

  3. Again, use the Biot-Savart law.

       dB = (µ0/4) I dl x r/r2

    Integrate dB around the semicircle. We note that I, µ0, 4 and r2=R2 are all constants (a circle is at a constant radius!). We also note that r and dl are perpendicular so their cross-product is easy. Thus the integral becomes

       B = (µ0/4) I/R2 dl

    The integral is thus the path length around the semicircle, i.e. R, and we have

       B = (µ0/4)( I R)/R2 = (µ0 I) /( 4 R)

    plugging in R = 5 cm and I = 2 A we get

       B = 1.26 x10-5 T

    Problem 3
    Test 3
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    last updated: April 26 1998