## Q. 5.7

A toroidal coil has N turns closely wound on an air core and carries a steady current I as shown in Fig. 5.10. The toroid has an inner radius a and an outer radius b. Find H everywhere.

## Verified Solution

The closely wound toroidal coil has cylindrical symmetry (rotational symmetry about the z-axis), and thus the resulting H is independent of Φ. If the toroidal coil is rotated about the x-axis by 180º , it appears the same except for the reversed direction of the current in the coil. Note that if we reverse the current in the original coil, the direction of the resultant H should also be reversed. In view of this, we see that H cannot have the ρcomponent, because the vector component $H_{\rho } \pmb{a}_{ \rho }$ appears the same even if it is rotated about the x-axis by 180º . From these symmetry considerations so far, we arrive at a conclusion that H is of the form

$\pmb{H}= H_{\phi } (\rho , z)\pmb{a}_{ \phi } + H_{z} [(\rho ,z)\pmb{a}_{z}$. Assuming an ideal toroidal coil in which no current penetrates the $\phi = \phi _{1}$ plane( $\phi _{1}$ is constant), we see that the line integral of H around any closed path $C _{1}$, which lies flat on the $\phi = \phi _{1}$ plane, is zero, i.e.,

$\oint_{C_{1}}{\pmb{H} \pmb{\cdot } d \pmb{l}} = \oint_{C_{1}}{(H_{\phi } \pmb{a}_{\phi } + H_{z} \pmb{a}_{z}) \pmb{\cdot } (d \rho \pmb{a}_{\rho } + d z \pmb{a}_{z})} = \oint_{C_{1}}{H_{z} (\rho , z) dz} \\ \quad \quad \quad \quad = 0$                                                                                                    (5-25)

Here, we use $d \pmb{l} = d \rho \pmb{a}_{\rho } + dz \pmb{a}_{z}$ in the $\phi =\phi _{1}$ plane. Suppose $C_{1}$ is a rectangle with the left side at $\rho =\rho _{1}$ and the right side at $\rho =\rho _{2}$, being parallel to the z-axis in the $\phi =\phi _{1}$ plane. Since Eq. (5-25) should be satisfied regardless of the height of $C_{1}$, we obtain the relation $H_{z} (\rho_{1} , z) = H_{z} (\rho_{2} , z)$ from Eq. (5-25), implying that $H_{z}$ is constant everywhere. The finite extent of the toroidal coil, however, assures that H = 0 at infinity, meaning that $H_{z} = 0$ everywhere. From the discussions so far, we conclude that H of the toroidal coil is of the form $\pmb{H} = H_{\phi} (\rho, z) \pmb{a}_{\phi}$.

Inside the toroidal coil (a < ρ < b), taking a circle of radius ρ as an Amperian path C, from Ampere’s circuital law we obtain

$\oint_{C}{\pmb{H} \pmb{\cdot } d \pmb{l}} = \oint_{C}{H_{\phi } \pmb{a}_{\phi } \pmb{\cdot } \rho d \phi \pmb{a}_{\phi } } =\rho H_{\phi } \int_{0}^{2\pi }{d \phi }= 2 \pi \rho H_{\phi }$

Since C encloses the total current of NI, Ampere’s circuital law gives

$H_{\phi } = \frac{NI}{2 \pi \rho }$

Inside the toroidal coil, we have

$\pmb{H} = H_{\phi } \pmb{a}_{\phi } = \frac{NI}{2 \pi \rho }\pmb{a}_{\phi }$                                                    (inside)                                           (5-26a)

Outside the toroidal coil, no net current is enclosed by C. Thus,

H = 0                                               (outside)                                                       (5-26b)

We notice that the final magnetic field intensity given in Eq. (5-26a) is of the form $\pmb{H} = H_{\phi } (\rho ) \pmb{a}_{\phi }$; it is not against our expectation of $\pmb{H} = H_{\phi } (\rho, z ) \pmb{a}_{\phi }$.