**Magnetic Field in a Solenoid Formula**

A solenoid is a long wire that has been twisted into the shape of a helix and is used to create a consistent magnetic field. If the turns are closer together, it can be perceived as a circular loop. The internal magnetic field of a solenoid becomes more homogeneous as its length rises. A solenoid’s total magnetic field equals the sum of magnetic fields produced at each of its turns. Because its length is far greater than the radius of the turns, an ideal solenoid has zero external fields and a uniform internal field.

A solenoid becomes an electromagnet when a current runs through it. Magnetic door locks, water-pressure valves in air conditioning systems, MRI machines, hard disc drives, speakers, microphones, and other electronic devices all use electromagnets.

The following are some of the most critical factors that influence the solenoid:

- In the solenoid, the number of rotations per unit length
- The strength of the current in the solenoid’s coil
- Inside the solenoid, there is a material.

**Solved Example:**

**Q1. Determine the magnetic field produced by a solenoid with a length of 80 cm and a coil with 360 turns and a current of 15 A.**

Number of turns N = 360

Current I = 15 A

Permeability μ_{0} = 1.26 × 10^{-6} T/m

Length L = 0.8 m

The magnetic field in a solenoid formula is given by,

B = μ_{0 }IN / L

B = (1.26× 10^{-6} × 15 × 360) / 0.8

B = 8.505 × 10^{-3}N/Amps m

The magnetic field generated by the solenoid is 8.505 × 10^{-3} N/Amps m

**Q2. The magnetic field of a solenoid with a diameter of 40 cm is 2.9×10 ^{-5} N/Amps m. Determine the current flowing through it if it has 300 turns.**

Number of turns N = 300

Length L = 0.4 m

Magnetic field B = 2.9 10^{-5} N/Amps m

The magnetic field formula is given by

B = μ_{0 }IN /L

The current flowing through the coil is expressed by

I = BL / μ_{0}N

I = (2.9 10^{-5} 0.4 )/ (1.26 × 10^{-6} × 300)

I = 30.6 mA.