12.6 Solenoids and toroids  (Page 3/5)

We begin by assuming cylindrical symmetry around the axis OO ’. Actually, this assumption is not precisely correct, for as part (b) of [link] shows, the view of the toroidal coil varies from point to point (for example, $P_1,P_2,$ and $P_3$ ) on a circular path centered around OO ’. However, if the toroid is tightly wound, all points on the circle become essentially equivalent [part (c) of [link] ], and cylindrical symmetry is an accurate approximation.

With this symmetry, the magnetic field must be tangent to and constant in magnitude along any circular path centered on OO ’. This allows us to write for each of the paths $D_1,D_2,$ and $D_3$ shown in part (d) of [link] ,

Ampère’s law relates this integral to the net current passing through any surface bounded by the path of integration. For a path that is external to the toroid, either no current passes through the enclosing surface (path $D_1$ ), or the current passing through the surface in one direction is exactly balanced by the current passing through it in the opposite direction $(\text{path}{D}_3).$ In either case, there is no net current passing through the surface, so

and

The turns of a toroid form a helix, rather than circular loops. As a result, there is a small field external to the coil; however, the derivation above holds if the coils were circular.

For a circular path within the toroid (path $D_2$ ), the current in the wire cuts the surface N times, resulting in a net current NI through the surface. We now find with Ampère’s law,

and

The magnetic field is directed in the counterclockwise direction for the windings shown. When the current in the coils is reversed, the direction of the magnetic field also reverses.

The magnetic field inside a toroid is not uniform, as it varies inversely with the distance r from the axis OO ’. However, if the central radius R (the radius midway between the inner and outer radii of the toroid) is much larger than the cross-sectional diameter of the coils r , the variation is fairly small, and the magnitude of the magnetic field may be calculated by [link] where $r=R.$

Summary

• The magnetic field strength inside a solenoid is
  
  $B={\mu }_{0}nI\text{(inside a solenoid)}$
  
  where n is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.
• The magnetic field strength inside a toroid is
  
  $B=\frac{{\mu }_{o}NI}{2\pi r}\text{(within the toroid)}$
  
  where N is the number of windings. The field inside a toroid is not uniform and varies with the distance as 1/ r .

Conceptual questions

Problems

A solenoid is 40 cm long, has a diameter of 3.0 cm, and is wound with 500 turns. If the current through the windings is 4.0 A, what is the magnetic field at a point on the axis of the solenoid that is (a) at the center of the solenoid, (b) 10.0 cm from one end of the solenoid, and (c) 5.0 cm from one end of the solenoid? (d) Compare these answers with the infinite-solenoid case.

Got questions? Get instant answers now!