### Solenoid Magnetic Field Calculator:

Find the amount of current flowing through the solenoid, the number of turns and the length of the solenoid.

Multiply the number of turns with the vacuum permeability constant and the current. Divide the product by the length of the solenoid to check the magnetic field.

A solenoid is a wire tightly wound into a long, thin coil. When current flows through a wire, a magnetic field surrounds it. The magnetic field outside an infinitely long solenoid is zero and the charge inside the solenoid is zero.

The formula for calculating the magnetic field inside the solenoid is along the lines. The Solenoid Magnetic Field Calculator helps calculate the magnetic field inside a long solenoid.

A solenoid is a wire tightly wound into a long, thin coil. When current flows through a wire, a magnetic field surrounds it.

The magnetic field outside an infinitely long solenoid is zero and the charge inside the solenoid is zero. The formula for calculating the magnetic field inside the solenoid is along the lines.

### Solenoid Magnetic Field Formula:

Solenoid Magnetic field formula is B_{(A)} in Ampere is equal to the vacuum permeability μ_{0(Tm/A)} in Tesla meter per Ampere and multiply the current flowing through a solenoid I_{(A)} in Ampere and multiply the number of turns N and divided by the length of the solenoid L_{(m)} in meter.Hence the solenoid magnetic field formula can be written as

B_{(A)} = μ_{0(Tm/A)} * I_{(A)} *N / L_{(m)}

Where,

B= is the magnetic field of the solenoid

I = is the current flowing through a solenoid

μ_{} = is vacuum permeability and its value is 1.26 *10^{-7} Tm/A

N= is the number of turns

L= is the length of the solenoid

### Example:1

A solenoid with 250 turns carries a current of 17A and the length is 89m.What is the magnetic field inside the coil?

### Answer:

Given that

Number of turns N=250

Current I =17A

Length L=89m

Magnetic field of the solenoid formula is

B_{(A)} = μ_{0(Tm/A)} * I_{(A)} *N / L_{(m)}

B= 1.25664 *10-6 *250 *17 /89

Magnetic field B =0.000060008089 T

### Example:2

A solenoid with 19 turns carries a current of 17A and the length is 25m.What is the magnetic field inside the coil?

### Answer:

Given that

Number of turns N=19

Current I =17A

Length L=25m

Magnetic field of the solenoid formula is

B_{(A)} = μ_{0(Tm/A)} * I_{(A)} *N / L_{(m)}

B= 1.25664 *10^{-6} *19 *17 /25

Magnetic field B =0.0000162357 T.