Physics Formulas

Magnetic Field Strength Formula

This article deals with the magnetic field strength formula. Magnetic Field Strength refers to one of two ways that the expression of a magnetic field can take place. It is certainly different from the magnetic flux density. Furthermore, the formation of a magnetic field takes place when a wire carries an electric current. The direction of the magnetic field is dependent on the direction of the current. The visualization of the magnetic field can take place as field lines. Also, the magnetic field strength definitely corresponds to the density of the field lines.

magnetic field strength formula

What is Magnetic Field Strength

Magnetic field strength refers to the ratio of the MMF which is required to create a certain Flux Density within a certain material per unit length of that material. Some experts also call is as the magnetic field intensity.

Furthermore, the magnetic flux refers to the total number of magnetic field lines that penetrate an area. Furthermore, the magnetic flux density tends to diminish with increasing distance from a straight current-carrying wire or a straight line which connects a pair of magnetic poles around which the magnetic field is stable.

Magnetic field strength refers to a physical quantity that is used as one of the basic measures of the intensity of the magnetic field. The unit of magnetic field strength happens to be ampere per meter or A/m.

Furthermore, the symbol of the magnetic field strength happens to be ‘H’. Magnetic field strength is a quantitative measure of strength or the weakness of the magnetic field. Also, it is the force which a unit north pole of one-weber strength experiences at a particular point in the magnetic field.

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Magnetic Field Strength Formula and Derivation

First of all, the formula for magnetic field magnitude is:

B = \(\frac{\mu _{0}I}{2\pi r}\)

B = magnetic field magnitude(Tesla,T)

\(\mu _{0}\) = permeability of free space \(4\mu \times 10^{-7}T.\frac{m}{A})\)

I = magnitude of the electric current( Ameperes,A)
r = distance(m)

Furthermore, an important relation is below

H = \(\frac{B}{\mu m}\)

H = \(\frac{B}{\mu _{0}}\) – M

The relationship for B can be written in this particular form

B = \(\mu _{0}\left ( H + M \right )\)

H and M would have the same units, amperes/meter. In order to further distinguish B from H, experts sometimes call the magnetic flux density or the magnetic induction. Furthermore, the quantity M in these relationships is the magnetization of the material.

Another form which is in common usage for the relationship between B and H is

B = \(\mu _{m}H\)


\(\mu\) = \(\mu _{m}\) = \(K_{m}\mu _{0}\)

Here, \(\mu _{0}\) is the magnetic permeability of space. \(K_{m}\) refers to the relative permeability of the material. Moreover, in case the material does not respond to the external magnetic field by the production of any magnetization, so \(K_{m}\) = 1. Another magnetic quantity which is in magnetic quantity is the magnetic susceptibility explains how much the relative permeability differs from one.

Magnetic susceptibility \(\chi _{m}\) = \(K_{m}\) – 1

The unit for the magnetic field strength which is H can be derived from its relationship to the magnetic field B. B = \(\mu H\). Moreover, the unit of magnetic permeability \(\mu\) happens to be \(\frac{N}{A^{2}}\). Therefore, the unit for the magnetic field strength is:

\(T\left ( \frac{N}{A^{2}} \right )\) = \(\frac{\frac{N}{Am}}{\frac{N}{A^{2}}}\) = A/m
Also, a unit for the magnetic field strength which is old is the oersted: 1A/m = 0.01257 oersted.

Solved Example on Magnetic Field Strength

Q1 Calculate the magnetic field strength inside a solenoid which is 2 m long and has 2000 loops. Furthermore, it carries a 1600 A current?

A1 In order to find the magnetic field strength inside the solenoid, one must use B = \(\mu _{0}Ni\). Furthermore, one must note the number of loops per unit length:

n = \(\frac{N}{\iota }\) = \(\frac{2000}{2}\) = 1000m-1 = 10 cm-1.

Now, one must substitute known values

B = \(\mu _{0}Ni\) = \(\left ( 4\pi 10^{-7}T.\frac{m}{A}\right )\left ( 1000m^{-1} \right )\left ( 1600A \right )\)
= 2.01 T.

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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