Page 345 - Intro to Tensor Calculus
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                                            Figure 2.6-6. Magnetic field around wire.

               Biot-Savart Law
                   The Biot-Savart law for magnetostatics describes the magnetic field at a point P due to a steady line
               current moving along a curve C and is

                                                              Z  ~
                                                    ~
                                                   B(P)=   µ 0   I × b e r  ds                        (2.6.59)
                                                           4π      r 2
                                                               C
               with units [N/amp · m] and where the integration is in the direction of the current flow. In the Biot-Savart
                                                                                                    ~
                                                          2
               law we have the constant µ 0 =4π × 10 −7  N/amp which is called the permeability of free space, I = I b e t is
               the current flowing in the direction of the unit tangent vector b e t to the curve C, b e r is a unit vector directed
               from a point on the curve C toward the point P and r is the distance from a point on the curve to the
                                                                                                 ~
               general point P. Note that for a steady current to exist along the curve the magnitude of I must be the
               same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface
                                            ~
                       ~
               currents K and volume currents J the Biot-Savart law is written
                                                               ZZ   ~
                                                      ~
                                                     B(P)=   µ 0   K × b e r  dσ
                                                             4π   S   r 2
                                                               ZZZ   ~
                                                      ~
                                              and    B(P)=   µ 0     J × b e r  dτ.
                                                             4π    V   r 2
               EXAMPLE 2.6-5.
                                                                                                        ~
                                             ~
                   Calculate the magnetic field B a distance h perpendicular to a wire carrying a constant current I.
               Solution: The magnetic field circles around the wire. For the geometry of the figure 2.6-6, the magnetic
               field points out of the page. We can write

                                                  ~
                                                  I × b e r = I b e t × b e r = Iˆ e sin α
               where ˆ e is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpen-
               dicularly.
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