Name: Right Hand Thumb Rule
Uploaded: 2019-07-19
Duration: 3 min 25 s
Description: When electricity (conventional current) flows in a long straight wire, it creates a circular or cylindrical magnetic field around the wire according to the r...

Question

Derive an expression for the magnetic field inside a toroid. Explain how it is different from the result obtained in case of a tightly wound long-solenoid.

Accepted Answer

Magnetic Field of a Toroidal (Endless) Solenoid : Toroidal 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 has been bent into a circular shape to close on itself . Let $n$ be the number of turns per unit length and it carries a current $I$ . Due to the current magnetic field is produced inside the toroid. Inside the toroid magnetic field lines are in the form of concentric rings. By symmetry, the direction of the magnetic field $\vec{B}$ at any point is tangential at each point of the circle and is same at all points on any such circle.

Magnetic Field inside the Core of Toroidal : Let there be a point $P$ inside the core of the toroidal. Consider a circle through the point $P$ lying within the space enclosed by the windings of the toroid. Let a be the radius of the circle. Applying Ampere’s law to this circle, we write

$\oint \vec{B} \cdot \vec{d l} = μ_{0} I$ ,

where $I$ is the net current enclosed by the circle. Now,

$\oint \vec{B} \cdot \vec{d 1} = \oint Bdl = B (2 πa)$

because $\vec{B}$ and $\vec{dl}$ are along the same direction at any point on the circle, and

$I = {NI}_{0}$

where $N$ is the total number of turns in the toroid. (Every turn passes once through the area enclosed by the circle through $P$ ). Therefore, eq. (i) becomes

$B (2 πa) = μ_{0} {NI}_{0}$

or $B (2 πa) = μ_{0} {NI}_{0}$

Thus, the field $B$ varies with $a$ . (In contrast to the solenoid, $B$ is not constant over the cross-section of the toroid).
If $n$ is the number of turns per unit length of the toroid, then $n = N / 2 π a$ a, and we have

$B = μ_{0} {nI}_{0}$

The field at a point such as $Q$ is zero because there is no current enclosed by the circle through $Q$ . Similarly, the field at a point such as $R$ is also zero, because each turn of the winding passes twice through the area enclosed by the circle through R, carrying equal currents in opposite directions, so that the net current enclosed by this circle is zero.
The field of a toroid is thus zero at all points except within the core.

Derive an expression for the magnetic field inside a toroid. Explain how it is different from the result obtained in case of a tightly wound long-solenoid.

Important Questions on Moving Charges and Magnetic Field

Learn and Prepare for any exam you want !