 |
| The Solenoid and the Toroid |
|
|
| Toroid is a hollow circular ring (like a medu vadai) on which a large number of turns of a wire are wound. |
| |
 |
| |
The above figure represents a toroid wound with a wire carrying a current I. Consider path 1, by symmetry , if there is any field at all in
this region, it will be tangent to the path at all point and 
will equal the product will equal the product of B and the circumference d = 2pr of the path. The current through the path however is zero and hence from Ampere's law the field B must be zero. |
| |
| Similarly, if there is any field at path 3, it will also be tangent to the path at all points. Each turn of the winding passes twice through the area bounded by this path, carrying equal currents in opposite directions. The net current though the area is therefore zero and hence B = 0 at all points of the path. |
| |
| The field of the toroidal solenoid is therefore confined wholly to the space enclosed by the windings. |
| |
| If we consider path 2, a circle of radius r, again by symmetry the field is tangent to the path and |
| |
 |
| |
| Each turn of the winding passes once through the area bounded by path 2 and total current through the area is NI, where N is the total number of turns in the windings. |
| |
| Using Ampere's law |
| |
 |
| |
 |
| |
| If the radial thickness of the core is small, field is almost constant across the section. |
| |
| Here 2pr circumferential length of to the toroid. |
| |
 |
| |
| Conclusion: |
| |
| Field outside the toroid and inside the core of the toroid is zero and within the toroid = m0ni |
| |
|
 |
| |
| The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the interior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. |
| |
| Solenoid is long wire wound in form of helix such that the length of solenoid is large compared to the radius of the closely spaced turns. |
| |
 |
| |
| In the diagram the upper dots represent the current coming out of the paper. Using the right hand rule the fields lines inside the solenoid go from left to right. Similarly the crosses on the lower side represent the current going into the paper. The field lines inside the solenoid also go from left to right. The two fields being in the same direction add up but outside the solenoid they cancel (i.e., the field contribution due to the dots and crosses). |
| |
| To find the magnetic field due to a solenoid consider the Amperian loop (imaginary closed path) as shown in the diagram. |
| |
The field along cd is zero as it is outside the solenoid. Along da and bc the transverse section the field is zero outside the solenoid (also, B is
perpendicular to dl so  = 0). Therefore the only
contribution is from ab. Let the length ab be 'h'. If there are n turns per unit length, then the enclosed current ie is |
| |
| ie = i (nh), where i is the current in the solenoid. |
| |
 |
| |
 |
| |
 |
| |
| |
| |
 |
| |
| direction is given by the right hand rule. |
| |
