Solenoid & Toroid

What is a solenoid?

A cylindrical coil of enameled ( i.e. insulated ) wire tightly wound in the form of a helix, with diameter of the coil smaller than its length, is called a solenoid.

A solenoid is consisting of a number of circular loops through which current flows. This current produces magnetic field around the coil.

Use of Solenoid

These are extremely useful as electromagnets. Their uses are –

Used as an element in door bells and calling bells.
It is used as paintball guns in painting work.
These are used in automated factory equipment in contactors, relays, actuators, timers etc.
Used in electrical contactors and circuit breakers for remote operation of high voltage circuits.
Also used in hydraulic valves and switches.
Used as induction coils or an impedance device.

Features of a Solenoid

A solenoid in which wire turns are tightly packed and their number is very large is termed as ideal solenoid.

Consider about a solenoid of coil diameter $( D )$ , wire diameter $( d )$ and length $( L )$ as shown in figure.

It has the following features –

The ratio of coil diameter to wire diameter $\left ( \frac {D}{d} \right )$ is kept more than $10$ .
Its length $( L )$ is kept more than its coil diameter $( D )$ .
Each turn of a solenoid is regarded as a perfect circular loop.
Total magnetic field due to the whole solenoid is equal to the vector sum of the magnetic field of each turn.
The magnetic field inside a long solenoid is uniform and strong magnetic field and outside it is non-uniform and weak magnetic field, which can be neglected.
The magnetic field of an ideal solenoid at a point outside the solenoid is practically zero.
Field lines are shown in figure below.

Magnetic Field of a Solenoid

Consider about a long solenoid of length $( l )$ , total number of turns $( N )$ and carrying a current $( I )$ .

Then, number of turns per unit length will be $n = \left ( \frac {N}{l} \right )$ .

Consider a point $P$ inside the solenoid. A rectangular Amperian loop $ABCD$ is passing through the point $P$ as shown in figure.

Then, line integral of magnetic field across the loop $ABCD$ will be –

$\oint \vec {B} \vec {dl} = \int\limits_{A}^{B} \vec {B} . \vec {dl} + \int\limits_{B}^{C} \vec {B} \vec {dl} + \int\limits_{C}^{D} \vec {B} \vec {dl} + \int\limits_{D}^{A} \vec {B} \vec {dl}$ ……… (1)

$( \vec {B} )$ is perpendicular to the paths $BC \ \text {and} \ AD$ . Therefore, for these paths, angle between $( \vec {B} )$ and $( \vec {dl} )$ is $( 90 \degree )$ .

Hence, $\quad \int\limits_{A}^{B} \vec {B} \vec {dl} = \int\limits_{C}^{D} \vec {B} \vec {dl}$ $= \int B {dl} \cos 90 \degree = 0$

Also, the path $CD$ is outside the solenoid and outside a solenoid the value of $( \vec {B} )$ can be neglected. So –

$\int\limits_{C}^{D} \vec {B} \vec {dl} = 0$

For path $AB$ the direction of $( \vec {B} )$ and $( \vec {dl} )$ is same i.e. $( \theta = 0 )$

Hence, equation (1) becomes –

$\oint \vec {B} \vec {dl} = \int\limits_{A}^{B} \vec {B} \vec {dl} = \int\limits_{A}^{B} B {dl} \cos 0 \degree$

Or, $\quad \oint \vec {B} \vec {dl} = \int\limits_{A}^{B} B {dl} = B \int\limits_{A}^{B} {dl}$ .

But, $\quad \int\limits_{A}^{B} {dl} = l$

Therefore, $\quad \oint \vec {B} \vec {dl} = B l$ ………. (2)

Now, according to Ampere’s circuital law –

$\oint \vec {B} \vec {dl} = \mu_0 \ \times \ \text {net current enclosed by the loop} ABCD$ .

Or, $\quad \oint \vec {B} \vec {dl} = \mu_0 \ \times \ \text {number of turns in the loop} ABCD \ \times \ I = \mu_0 n l I$ ………. (3)

Comparing equations (2) and (3), we get –

$B l = \mu_0 n l I$

Or, $\quad B = \mu_0 n I$

Thus, magnetic field within an infinitely long solenoid is given by –

$B = \mu_0 n I = \mu_0 \left ( \frac {N}{l} \right ) \times I$

$= \left ( \frac {\mu_0 N I}{l} \right )$ ……… (4)

Variation of Magnetic Field of a Solenoid

Magnetic field of a long solenoid vary as –

Inside a long solenoid magnetic field is uniform and strong magnetic field.
The magnetic field at a point outside the solenoid is weak, non uniform and negligible.
Magnetic field at each end reduces to the half of the value inside the solenoid.
The variation of the magnetic field with distance from centre of the solenoid and along the axis of solenoid is shown in the figure.

Therefore, it reveals that the magnetic field out side a solenoid is not important however the magnetic field inside a solenoid is of our great importance for use. This magnetic field inside a solenoid can be increased by –

Inserting a magnetic material like iron rod of relative permeability $\left ( \mu_r \right )$
By increasing the number of turns of the coil.
By increasing the current flowing through the coil.

Toroid

A toroid is a hollow and thick circular ring on which a large number of turns of an insulated wire are closely wound.

A toroid can be considered as a ring shaped closed solenoid as shown in figure.

Magnetic Field of a Toroid

Consider a toroid having $( n )$ number of turns per unit length. Let, $( I )$ is the current flowing through the toroid.

The magnetic field lines mainly remain in the core of a toroid in the form of concentric circles.
Outside and inside of the core of toroid, magnetic field is negligible.

Consider about an imaginary Amperian loop in form of a circle of radius $( r )$ as shown in figure. For this circular loop, we have the relation –

$\oint \vec {B} \vec {dl} = \oint B {dl} \cos \theta$

By symmetry, magnetic field $B$ in the coil is constant and is along the tangent to path $( \vec {dl} )$ . So, angle $( \theta )$ between $( \vec {B} )$ and $( \vec {dl} )$ is zero.

Hence, $\quad \oint \vec {B} \vec {dl} = \oint B {dl} \cos 0 \degree = B \oint dl$

$= B \ \times \ \text {Circumference of the circle of radius} ( r )$

$= B \times ( 2 \pi r )$ ……….. (1)

According to Ampere’s circuital law –

$\oint \vec {B} \vec {dl} = \mu_0 \ \times \ \text {Net current enclosed by the circle of radius} ( r )$

$= \mu_0 \ \times \ \text {Total number of turns} \ \times \ I$ $= \mu_0 \left ( n \times 2 \pi r \right ) I$ …….. (2)

Comparing equations (1) and (2), we get –

$B \ \times \ ( 2 \pi r ) = \mu_0 \left ( n \times 2 \pi r \right ) I$

Or, $\quad B = \mu_0 n I$ ……. (3)

This equation gives the magnetic field produced by a toroid.

Total Magnetic Field of Toroid

If, $( N )$ is the total number of turns of a toroid, then –

$N = n \times ( 2 \pi r )$

Or, $\quad n = \left ( \frac {N}{2 \pi r} \right )$

Therefore, equation (3) can be written as –

$B = \left ( \frac {\mu_0 N I}{2 \pi r} \right )$ ……….. (4)

Variation of Magnetic Field of Toroid

Magnetic field of a toroid vary as –

For any point inside the empty space surrounded by toroid as well as outside the toroid magnetic field $( B )$ is zero.

Inside the core of toroid magnetic field is uniform and strong magnetic field.
The magnetic field at a point outside the toroid is weak, non uniform and negligible.
Magnetic field inside the empty space surrounded by toroid is also zero.

See numerical problems based on this article.