The solenoid and the toroid are two pieces of equipment which generate magnetic fields. The television uses the solenoid to generate magnetic fields needed. The synchrotron uses a combination of both to generate the high magnetic fields required. In both, solenoid and toroid, we come across a situation of high symmetry where Ampere’* law can be conveniently applied.
(1) The solenoid : Consider a long solenoid where the solenoid’* length is large compared to its radius. It consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced. So each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enamelled wires are used for winding so that turns are insulated from each other.
(a) The magnetic field due to a section of the solenoid.
(b) The magnetic field of a finite solenoid.
Fig. displays the magnetic field lines for a finite solenoid. We show a section of this solenoid in an enlarged manner in Fig.(a). Figure (b) shows the entire finite solenoid with its magnetic field. In Fig.(a), it is clear from the circular loops that the field between two neighbouring turns vanishes. In Fig. (b)we see that the field at the interior mid-point P is uniform, strong and along the axis of the solenoid. The field at the exterior mid-point Q is weak and moreover is along the axis of the solenoid with no perpendicular or normal component. As the solenoid is made longer it appears like a long cylindrical metal sheet.
Figure represents this idealized picture. The field outside the solenoid approaches zero. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis.
Consider a rectangular Amperian loop abcd. Along cd the field is zero as argued above. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h.
Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’* circuital law [Eq. (b)] BL = µ0Ie, B h = µ0I (n h)
B = µ0 n I ---------- (1)
The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field. We shall see in the next chapter that a large field is possible by inserting a soft iron core inside the solenoid.
(2) The toroid :
The toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a solenoid which has been bent into a circular shape to close on itself. It is shown in Fig. (a) carrying a current I. We shall see that the magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero. The field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns.
Figure (b) shows a sectional view of the toroid. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines. By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop. The circular areas bounded by loops 2 and 3 both cut the toroid: so that each turn of current carrying wire is cut once by the loop 2 and twice by the loop 3.
Let the magnetic field along loop l be B1 in magnitude.
Then in Ampere’* circuital law [Eq.(a)], L = 2π r1.
However, the loop encloses no current, so Ie = 0.
Thus, B1 (2 π r1) = µ0(0), B1 = 0 Thus, the magnetic field at any point P in the open space inside the toroid is zero.
We shall now show that magnetic field at Q is likewise zero. Let the magnetic field along loop 3 be B3. Once again from Ampere’* law L = 2 π r3. However, from the sectional cut, we see that the current coming out of the plane of the paper is cancelled exactly by the current going into it.
Thus, Ie = 0, and B3 = 0. Let the magnetic field inside the solenoid be B. We shall now consider the magnetic field at . Once again we employ Ampere’ law in the form of Eq. (a). We find, L = 2πr.
The current enclosed Ie is (for N turns of toroidal coil) N I.
B (2πr) = µ0NI
We shall now compare the two results: for a toroid and solenoid. We re-express Eq. (2) to make the comparison easier with the solenoid result given in Eq. (1). Let r be the average radius of the toroid and n be the number of turns per unit length. Then N = 2πr n = (average) perimeter of the toroid × number of turns per unit length and thus
B = µ0 n I, -------- (3)
i.e., the result for the solenoid! In an ideal toroid the coils are circular. In reality the turns of the toroidal coil form a helix and there is always a small magnetic field external to the toroid.
