# Magnetic Field In A Solenoid

A solenoid is the arrangement of the coil in such a way that it generates a magnetic field and they behave as a bar magnet. A solenoid is made by wrapping the electric wire over a soft iron core. Now we know that a solenoid behaves as a bar magnet so it has a magnetic field all around it. And the magnetic field of the solenoid when the current passes through it is studied further below.

In this article, we will learn about, Solenoids, the magnetic field inside it, the formula to calculate the strength of the magnetic field, and others in detail.

## Solenoid Definition

A solenoid is a simple device that can be made by wrapping metal wires such as copper wires around a metal core. In other words, a solenoid is a long wire twisted in the shape of a helix around a metal core. When both ends of a solenoid are connected to a power supply, it generates a uniform magnetic field.

When an electric current flows through the coil, it generates a uniform magnetic field. This allows small solenoids to perform powerful work, such as slamming shut a valve. The ability of a solenoid to convert electricity to magnetism and vice versa makes it a versatile component in various applications.

## Magnetic Field Inside a Solenoid

A solenoid is a long wire twisted in the shape of a helix that aids in generating a uniform magnetic field. It can be interpreted as a circular loop if the turns have less space between them. The uniformity of the internal magnetic field of a solenoid increases with an increase in its length. The total magnetic field of a solenoid is equal to the summation of magnetic fields produced at each of its turns. An ideal solenoid has zero external fields and a uniform internal field as its length is way more than the radius of the turns.

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## Magnetic Field in a Solenoid Formula

Magnetic Field in Solenoid depends on various factors such as the number of turns per unit length, the current strength in the coil, and the permeability of the material used in the solenoid. The magnetic field of a solenoid is given by the formula:

B = Î¼oIN/L

where,

• Î¼o is the permeability constant with a value of 1.26 Ã— 10âˆ’6 T/m,
• N is the number of turns in the solenoid,
• I is the current passing through the coil,
• L is the coil length.

Note: The magnetic field in a solenoid is maximum when the length of the solenoid is greater than the radius of its loops.

## Derivation of Formula for Magnetic Field in Solenoid

The formula for the magnetic field inside a solenoid can be derived through Ampereâ€™s Circuital Law.

Consider a solenoid of length n such that current I pass through each turn of the solenoid

We know the integral over a closed path rectangular in shape is given by,

Â  . . . (1)

Now the Total Current that passes through the solenoid is given by,

IT = Number of Turns in Solenoid Ã— Current Passing through Each Turn

â‡’ IT = (nh) (I) . . . (2)

Using Ampere’s law, we can conclude that,

From (1) and (2), we have

Bh = Î¼oITÂ

â‡’ Bh = Î¼o(nh)(l)

â‡’ B = Î¼onIÂ

Putting n = N/L as n is the number of units per unit length, we get

B = Î¼oIN/L

This derives the formula for the magnetic field inside a solenoid.

## Applications of Solenoid

As solenoids exhibit an immediate reaction when electricity is applied, making them useful in resolving various applications. Some of these applications are as follows:

• Solenoids are used in various applications, including medical, locking systems, industrial use, automotive, and more.
• Electromagnets, Transformers, and Inductors are also applications of Solenoid which further has plenty of applications in real-world scenarios.
• Solenoids can be used to control valves electrically by applying mechanical force to the valve which can be used in door-locking systems to provide a secure closure.
• Solenoids are used in various appliances and products, including computer printers and fuel injection gear in cars.

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## Sample Problems on Magnetic Field in a Solenoid

Problem 1: Find the value of the magnetic field inside a solenoid of 2 m and 100 turns per unit length if 5A of current is passing through it.Â

Solution:

Given:

• n = 200
• L = 2
• I = 5

Find the number of turns using the formula n = N/L.

N = nL

â‡’ N = 200 (2)

â‡’ N = 400 turns

Using the formula for magnetic field we have,

B = Î¼oIN/L

â‡’ B = 4Ï€ Ã— 10â€“7 Ã— (400/2) Ã— 5

â‡’ B = 4Ï€ Ã— 10â€“7 Ã— 200 Ã— 5

â‡’ B = 12.56 Ã— 10-4 T

Problem 2: Find the value of the magnetic field inside a solenoid of 5 m and 500 turns per unit length if 10A of current is passing through it.

Solution:

Given:

• n = 500
• L = 5
• I = 10

Find the number of turns using the formula n = N/L.

N = nl

= 500 (5)

= 2500 turns

Using the formula for magnetic field we have,

B = Î¼oIN/L

= 4Ï€ Ã— 10â€“7 Ã— (2500/5) Ã— 10

= 4Ï€ Ã— 10â€“7 Ã— 500 Ã— 10

= 6.3 Ã— 10-2 T

Problem 3: Find the value of the number of turns for a solenoid of 4 m if 2A of current is passing through it and the field is 1.25 Ã— 10-4 T.

Solution:

Given:

• B = 1.25 Ã— 10-4
• L = 4
• I = 2

Using the formula for magnetic field we have,

B = Î¼oIN/L

â‡’ Â 1.25 Ã— 10-4 = 4Ï€ Ã— 10â€“7 Ã— (N/4) Ã— 2

â‡’ Â N = (1.25 Ã— 10-4)/(0.5 Ã— 1.26 Ã— 10âˆ’6)

â‡’ Â N = 100 turns

Problem 4: Find the value of the number of turns for a solenoid of 8 m if 6A of current is passing through it and the field is 1.88 Ã— 10-4 T.

Solution:

Given:

• B = 1.88 Ã— 10-4
• L = 8
• I = 6

Using the formula for magnetic field we have,

B = Î¼oIN/L

â‡’ 1.88 Ã— 10-4 = 4Ï€ Ã— 10â€“7 Ã— (N/8) Ã— 6

â‡’ N = (1.88 Ã— 10-4)/(0.75 Ã— 1.26 Ã— 10âˆ’6)

â‡’ N = 200 turns

Problem 5: The line integral around a solenoid is 4 Ã— 10â€“7 T/m. Find the net current of the solenoid.

Solution:

We have,Â .

We know that,

IT = (4 Ã— 10â€“7)/(4Ï€ Ã— 10â€“7)

â‡’ IT = 0.31 A

## FAQs on Magnetic Field in a Solenoid

### Q1: What is Solenoid?

Answer:

Solenoid is a long wire wrapped around a metal core in a helix pattern and whenever solenoid is connected to a power source it generates a uniform magnetic field.

### Q2: What is Magnetic Field inside a Solenoid?

Answer:

The magnetic field inside a solenoid is relatively uniform and points along the axis of the solenoid.

### Q3: How is the Magnetic Field inside a Solenoid Calculated?

Answer:

The magnetic field inside a solenoid is uniform and can be calculated using the following formula:

B = Î¼oIN/L

Where,

• Î¼o is the permeability constant with a value of 1.26 Ã— 10âˆ’6 T/m,
• N is the number of turns in the solenoid,
• I is the current passing through the coil,
• L is the coil length.

### Q4: What is the Direction of Magnetic Field in a Solenoid?

Answer:

The direction of the magnetic field inside a solenoid is along the axis of the solenoid, and it points from the south pole to the north pole of the magnetic field.

### Q5: Where is Magnetic Field Maximum in Solenoid?

Answer:

The magnetic field in a solenoid is maximum when the length of the solenoid is greater than the radius of its loops.

### Q6: What is the relation between Magnetic Field inside a Solenoid and Current Flowing through it?

Answer:

The magnetic field inside a solenoid is directly proportional to the current flowing through it i.e., as the current increases, the magnetic field strength inside the solenoid also increases.

### Q7: What is Magnetic Field in a Solenoid Equation?

Answer:

The magnetic field in a solenoid equation is,

B = Î¼oIN/L

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