Class 12 > Unit # 18:Magnetic Fields >Magnetic Field due to Toroid

Talha's Physics Academy

Electromagnetism - Magnetic Field of a Toroid

Using Ampere’s Law derive an expression for the magnetic field of induction due to a Toroid.

TOROID

"A Toroid or a circular solenoid is a coil of insulated copper wire wound on a circular core with close turns. When current passes through the toroid, a magnetic field is produced which is strong enough inside it, while outside the toroid the field is almost zero."

Figure: Toroid with closely packed windings and concentric circular paths for Ampere's law analysis.

MAGNETIC FIELD OF INDUCTION DUE TO A TOROID

Consider a toroid consisting of $N$ closely packed turns and carrying a current $I$. Imagine a circular curve of radius $r$ concentric with the core as shown in the figure. It is evident from symmetry that the magnetic field at all points of the curve must have the same magnitude and should be tangential to the curve at all points.

Evaluating the line integral of the magnetic field along this circular path:

$\oint \vec{B} \cdot d\vec{l} = \oint B \, dl \cos(0^\circ) = B \oint dl$

Since $\oint dl = 2\pi r$ (the circumference of the circular path of radius $r$), we get:

$\oint \vec{B} \cdot d\vec{l} = B (2\pi r) \quad \text{--- (i)}$

By Ampere’s Circuital Law:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 \times (\text{Total Current Enclosed})$

If the circular path (marked 2) is within the core, the area bounded by the curve will be threaded by $N$ turns, each carrying a current $I$. Thus, the total enclosed current is $NI$:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 (N I) \quad \text{--- (ii)}$

By comparing equation (i) and equation (ii), we get:

$B (2\pi r) = \mu_0 N I$

$B = \frac{\mu_0 N I}{2\pi r}$

This expression gives the magnetic field of induction inside the core of a toroid.

CASES

1. Inner Region (Path 1): If the circular path (marked 1) is outside the core on the inner side of the toroid, it encloses no current ($I_{\text{enclosed}} = 0$).
   Thus:
   $B (2\pi r) = \mu_0 (0) \implies B = 0$
2. Outer Region (Path 3): If the circular path (marked 3) is outside the core on the outer side of the toroid, each turn of the winding passes twice through the area bounded by this path, carrying equal currents in opposite directions. Thus, the net current through the area is zero ($I_{\text{enclosed}} = 0$).
   Thus:
   $B (2\pi r) = \mu_0 (0) \implies B = 0$

© 2026 Talha's Physics Academy. All rights reserved.