__Solenoid Current Calculator:__

__Solenoid Current Calculator:__

Enter the values of magnetic flux density, B_{(T)}, length of the solenoid, L_{(m)}, permeability of the material, μ_{(H/m)} and total number of windings, N to determine the value of Solenoid current, I_{s(A)}.

__Solenoid Current Formula:__

__Solenoid Current Formula:__

Solenoid current is the electric current that flows through the wire windings of a solenoid, which is a coil of wire designed to generate a magnetic field when an electric current passes through it.

The magnetic fields generated by each turn of the wire combine to create a strong and uniform magnetic field inside the solenoid. This field is concentrated along the axis of the coil and is much stronger inside the coil than outside.

The core material inside the solenoid (which can be air or a ferromagnetic material like iron) affects the strength of the magnetic field. The permeability μ of the core material plays a crucial role.

A ferromagnetic core significantly increases the magnetic field strength compared to an air core.

The number of turns, N of the wire and the length, L of the solenoid affect the magnetic field strength. More turns and a longer solenoid result in a stronger magnetic field.

Solenoid current, I_{s(A)} in amperes is calculated by dividing the product of magnetic flux density, B_{(T)} in Tesla, length of the solenoid, L_{(m)} in metres by the product of permeability of the material, μ_{(H/m)} in Henry per metres and total number of windings, N.

Solenoid current, I_{s(A)} = B_{(T)} * L_{(m)} / μ_{(H/m)} * N.

I_{s(A)} = solenoid current in amperes, A.

B_{(T)} = magnetic flux density in Tesla, T.

L_{(m)} = length of the solenoid in metres, m.

μ_{(H/m)} = permeability of the material in Henry per metres, H/m.

N = total number of windings.

__Solenoid Current Calculation:__

__Solenoid Current Calculation:__

- Calculate the solenoid current in a solenoid with a magnetic flux density of 0.01 Tesla, a length of 0.5 meters, a permeability of 12.56 * 10
^{-7}H/m, and 200 turns.

Given: B_{(T)} = 0.01T, L_{(m)} = 0.5m, μ_{(H/m)} = 4 * π * 10^{-7} H/m, N = 200.

Solenoid current, I_{s(A)} = B_{(T)} * L_{(m)} / μ_{(H/m)} * N.

I_{s(A)} = 0.01 * 0.5 / 12.56 * 10^{-7} * 200

I_{s(A)} = 0.005 / 2.5132 * 10^{-4}

I_{s(A)} = 19.9A.

- Given a solenoid with a magnetic flux density of 0.02 teslas, a length of 1 meter, a permeability of 12.56 * 10
^{-7}H/m, and a current of 15 amperes, calculate the number of turns.

Given: B_{(T)} = 0.02T, L_{(m)} = 1m, μ_{(H/m)} = 12.56 * 10^{-7} H/m, I_{s(A)} = 15A.

Solenoid current, I_{s(A)} = B_{(T)} * L_{(m)} / μ_{(H/m)} * N.

N = B_{(T)} * L_{(m)} / μ_{(H/m)} * I_{s(A)}

N = 0.02 * 1 / 12.56 * 10^{-7} * 15

N = 0.02 / 1.8849 * 10^{-5}

N = 1061.

__Solenoid Basics Explained – Working Principle:__

__Solenoid Basics Explained – Working Principle:__