Solenoids

1.0What is a Solenoid? 

A solenoid is essentially a long, tightly wound coil of wire that has a uniform magnetic field. It has a significant function in electrical and magnetic devices, including electromagnets, electric motors, and magnetic locks. The solenoid magnetic field produced by the coil is crucial in explaining how it works and, therefore, how other devices work.

2.0Structure of a Solenoid

A solenoid is usually a cylindrical coil of wire. It consists of several turns or loops of conducting wire wound in the shape of a helix. The wire is usually made of copper or other conductive material. When an electric current flows through the coil, it generates a magnetic field around it. The solenoid diagram below shows a simple solenoid coil. The magnetic field produced within a solenoid is uniform like it is given in this solenoid diagram, so solenoids are highly useful as tools for generating controlled environments of magnetism.

3.0Working Principle of a Solenoid

When a current is passed through the wire of a solenoid, it develops a magnetic field that surrounds the coil. Its direction is determined by the direction of the current itself and is found using the right-hand thumb rule. According to this rule, if you hold the solenoid such that your right-hand fingers curl in the direction of the current with it, then your left-hand thumb will point in the direction of the magnetic field.

4.0Types of Solenoids

There are different types of solenoids. Some of the most common are: 

Types	Solenoid Diagram
Air-Core Solenoid: Absent ferromagnetic Core with a weaker magnetic field. Example: Small electromagnets, basic experiments.
Iron-Core Solenoid: Has a ferromagnetic core (normally iron) with a strong magnetic field.  Examples: Electromagnets, MRI machines, and electric bells.
Rotary Solenoid: It creates rotational motion in place of linear motion(plunger moves)  Examples: Locks and switches requiring rotational motion.
Electromagnetic Solenoid: A magnetic field that is produced by electric current.  Examples: Relays, solenoid valves, and actuators.
Linear Solenoid: It creates linear motion in which the plunger moves.  Example: automatic doors and door locking systems.

5.0Magnetic Field of a Solenoid

The magnetic field (represented by B) of a solenoid is similar to that of a bar magnet; however, it is much stronger and more concentrated within a solenoid. The formula gives the magnetic field strength (B) inside a solenoid:

$B={\mu }_{0}nI$

Where:

• B is the magnetic field strength present inside the Solenoid,

The magnetic field strength at the ends is weaker than inside the solenoid and usually decreases with distance from the solenoid. The field is no longer uniform, and the field lines are spread out, which makes the calculation more complicated.

• $n=\frac{N}{L}$(Number of turns per unit length of the solenoid. Here, N is the number of turns in the Solenoid, and L is the length of the Solenoid)
• ${\mu }_0$ is the permeability of free space $(4\pi \times 10^{-7}Tm/A)$
• 𝐼 is the current passing through the Solenoid.

6.0Factors Affecting the Magnetic Field of a Solenoid

The strength of the magnetic field inside a solenoid can be influenced by several factors:

• Number of Turns (N): More turns result in a stronger magnetic field as the magnetic field from each turn adds up.
• Present (I): A greater current results in a stronger magnetic field.
• The Length of the Solenoid: A smaller solenoid produces a stronger magnetic field because its field lines are more closely packed.
• Permeability of the Core: The magnetic field strength is substantially increased if a ferromagnetic core exists for the solenoid because ferromagnetic materials tend to increase the magnetic field.

7.0Solved Examples

Q1. A solenoid is required to produce a magnetic field of 0.5T. If the current through the solenoid is 3 A and the length of the solenoid is 0.5 m, calculate the number of turns required for the solenoid.

Given:

Magnetic field (B)=0.5 T;

Current, I=3A

Length of solenoid, L=0.5m

Permeability of free space, $(4\pi \times 10^{-7}Tm/A)$

Solution:

Using the formula for the B, i.e., Magnetic field in a solenoid:

$B={\mu }_0\frac{N}{L}I$

Rearranging for N:

$N=\frac{BL}{{\mu }_{0}I}$

Substitute the values:

$N=\frac{0.5\times 0.5}{4\pi \times 10^{-7}\times 3}$

$N=\frac{0.25}{4\pi \times 10^{-7}\times 3}$

$N=\frac{0.25}{3.77\times 10^{-6}}\approx 66344$

Q2. A solenoid with 1200 turns and a length of 0.5 m produces a magnetic field strength of 0.3 T. Calculate the current flowing through the solenoid.

Given:

Number of turns, N=1200

Length of solenoid, L=0.5m

Magnetic field strength, B=0.3 T

Permeability of free space, ${\mu }_0=(4\pi \times 10^{-7}Tm/A)$

Solution:

We can use the formula for the magnetic field inside a solenoid:

$B={\mu }_0\frac{N}{L}I$

Rearranging to find 𝐼 (current):

$I=\frac{BL}{{\mu }_{0}N}$

Substitute the given values:

$I=\frac{0.3\times 0.5}{(4\pi \times 10^{-7})\times 1200}$

$I=\frac{0.15}{1.507\times 10^{-3}}\approx 99.6A$

Thus, the current flowing through the solenoid is approximately 99.6 A.

Problem 3: A long solenoid has 2000 turns and a length of 1 meter. The current passing through the solenoid is increasing from 0 A to 3 A in 0.5 seconds. The cross-sectional area of the solenoid is 2 cm².

1. Find the magnetic field inside the solenoid at the final moment when the current is 3 A.
2. Calculate the induced emf in the solenoid during the first 0.5 seconds.

Solution: 

1. The magnetic field inside the Solenoid 

$B={\mu }_0\frac{N}{L}I$

Here, ${\mu }_0=(4\pi \times 10^{-7}Tm/A)$ and $I_2=3A$

$B=(4\pi \times 10^{-7})\frac{2000}{1}3$

$B=4\pi \times 10^{-7}\times 2000\times 3$

$B=4\times 3.1416\times 10^{-7}\times 2000\times 3$

$B=4\times 3.1416\times 10^{-7}\times 6000$

$B=7.5398\times 10^{-3}T=7.54mT$

2. Induced EMF in the Solenoid 

EMF $\epsilon =-N\frac{d\varphi }{dt}$

Magnetic flux $\varphi =BA$

$\Delta \varphi =({\varphi }_{final}-{\varphi }_{initial}=(B_{final}A)-(BinitialA)$

In the start the current I₁ = 0, so the initial magnetic field $B_{initial}=0.$

$\Delta \varphi =(B_{final}A)=7.54\times 10^{-3}\times 2\times 10^{-4}$

$\Delta \varphi =1.508\times 10^{-6}Wb$

$\epsilon =-2000\times \frac{1.508\times 10^{-6}}{0.5}=-2000\times 3.016\times 10^{-6}$

$\epsilon =-6.032\times 10^{-3}V=-6.03mV$