Bar Magnet as an Equivalent Solenoid

A coil of wire is known as a solenoid. The solenoid is used in an electromagnet that can convert electrical energy into mechanical energy. A magnetic field is created by the solenoid and helps to create linear motion.  

In automobiles, solenoids are used to give the power to start. It works like a sprinkler system and has the shape of a valve. 

Electromagnets serve as great equipment compared to permanent magnets. We can regulate the magnetic field by switching on/off the electric current. This helps the electromagnets to be used as switches. 

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What is a Solenoid and How Does it Work?

The solenoid is a coil of wire. It is covered around the piston and is corkscrew in shape. A magnetic field develops in the solenoid when the electric current goes inside the wire. 

Solenoids with an activated magnetic field have positive and negative poles like other magnets. The negative end of the magnetic solenoid attracts, and the positive end repels. The electromagnetic field present inside the solenoid flows in the forward and backward motion of the piston.

What is the Use of Solenoid?

• Solenoid helps in the ignition system of an automobile. 

• The solenoid acts as a relay that draws metal close to each other. 

• The versatile nature solenoid makes it extremely useful. 

The application of solenoid is widely available, like:

• In doorbells

• Some automated factory equipment use solenoid to paintball guns.

• In a chime doorbell.  

What is a Bar Magnet?

We are familiar with the term bar magnet since our childhood. This rod or bar magnet attracts small iron pieces, or tiny copper wires, or anything small but made of metal. When we try to draw them closer to each other, they either repel or attract each other. 

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Magnets have two groups known as natural magnets and artificial magnets. The artificial magnets are the magnets created by man. Let’s discuss these magnets:

1. Natural Magnets: These magnets are found in nature and possess a weak magnetic field. A common example of natural magnets is Lodestones

2. Artificial Magnets: It is designed by men. These magnets create a stronger magnetic field. We can give any required designs to these magnets. When it is in the shape of a bar, then we call it a bar magnet.

Bar Magnet as an Equivalent Solenoid

We can demonstrate the bar magnet as a solenoid with the help of a current-carrying solenoid. To prove this, we need to calculate the axial field generated from the current-carrying solenoid.

Let’s visualize the figure given below and know about the solenoid:

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(Fig.1)

Here, a = radius of the solenoid

2l = length

n = number of turns per unit length

I = amount of current passing through the solenoid

Derivation of Bar Magnet as an Equivalent Solenoid

From Fig.1, let’s assume that dx is the small thickness of an element of the solenoid. dx is taken from point O at a distance of x. 

Here, OP = r 

Let’s find out the axis of the solenoid due to n turns:

dB = \[\frac{μ_{0}ndxIa^{2}}{2[(r-x)^{2}+a^{2}]^{\frac{3}{2}}}\]

Now, we should integrate from -I to +I then to get the total field magnitude:

B = \[\frac{μ_{0}nIa^{2}}{2}\int_{-I}^{I}\frac{dx}{[(r-x)^{2}+a^{2}]^{\frac{3}{2}}}[(r-x)^2]^{\frac{3}{2}}\] = r3

Here, B = \[\frac{μ_{0}nIa^{2}}{2r^{3}}\int_{-I}^{I}dx\] = \[\frac{μ_{0}nI}{2}\frac{2Ia^{2}}{r^{3}}\]

Finally, B =  \[\frac{μ_{0}}{4π}\frac{2m}{r^{3}}\]

The expression obtained above helps to understand as the magnetic moment of a bar is identical to the solenoid’s magnetic moment.

Difference Between Bar Magnet and Solenoid

Similarities Between A Bar Magnet and A Solenoid

Here are some similarities between a bar magnet and a solenoid:

1. Both of the magnets possess attractive and directive properties. 

2. Both of them can align themselves along with the external magnetic field.

3. They have the same magnetic field at the axial point.

4. Also, they have the same magnetic moment.

Solved Problem

A solenoid has a length of 80 cm rounded with the coils of 360 number of turns. The current of 15 A is allowed to pass through it. Calculate the produced magnetic field.

Ans: As per the question:

n = Number of turns of the coil = 360

μo = Permeability = 1.26 × 10−7 T/m

I = Current available = 15 A

L = Length = 0.8 m

The solenoid’s magnetic field formula is:

Magnetic field, B = INμo / L

B = (15 × 360 ×1.26 × 10−7) / 0.8

B = 8.505 × 10−4 N/Amp-m