Magnetic Field In A Solenoid Formula

Magnetic Field In a Solenoid

A coil of wire which is designed to generate a strong magnetic field within the coil is called a solenoid. Wrapping the same wire many times around a cylinder creates a strong magnetic field when an electric current is passed through it. N denotes the number of turns the solenoid has. More the number of loops, stronger is the magnetic field.

A solenoid is a type of electromagnet whose intention is to produce a controlled magnetic field. If the purpose of a solenoid is to impede changes in the electric current, it can be more specifically classified as an inductor. Â

The formula for the magnetic field of a solenoid is given by,

B = Î¼oIN / L

Where,

N = number of turns in the solenoid

I = current in the coil

L = length of the coil.

Please note that the magnetic field in the coil is proportional to the applied current and number of turns per unit length.

Solved Examples

Example 1

Determine the magnetic field produced by the solenoid of length 80 cm under the number of turns of the coil is 360 and the current passing through is 15 A.

Solution:

Given:

Number of turns N = 360

Current I = 15 A

Permeability Î¼o = 1.26 Ã— 10âˆ’6 T/m

Length L = 0.8 m

The magnetic field in a solenoid formula is given by,

B = Î¼oIN / L

B = (1.26Ã—10âˆ’6 Ã— 15 Ã— 360) / 0.8

B = 8.505 Ã— 10âˆ’3 N/Amps m

The magnetic field generated by the solenoid is 8.505 Ã— 10âˆ’4 N/Amps m.

Example 2

A solenoid of diameter 40 cm has a magnetic field of 2.9 Ã— 10âˆ’5 N/Amps m. If it has 300 turns, determine the current flowing through it.

Solution:

Given:

No of turns N = 300

Length L = 0.4 m

Magnetic field B = 2.9 Ã— 10âˆ’5 N/Amps m

The magnetic field formula is given by

B = Î¼oIN /L

The current flowing through the coil is expressed by

I = BL / Î¼oN

I = (2.9Ã—10âˆ’5 x 0.4 )/ (1.26Ã—10âˆ’6 Ã— 300)

I = 30.6 mA

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