## Magnetic Field Strength Formula :

Magnetic Field strength Formula is magnetization of the magnitude B_{(A/m)} in Ampere per meter is equal to product of permeability of free space μ_{0(m/A)} in meter per Ampere and multiply the magnitude of the electric current I _{(A)} in Ampere and divided by the two pi 2 π and multiply the distance r_{(m)} in meter. Hence the Magnetic Field Strength Formula can be written as

**B _{(A/m)}=μ_{0(m/A)}*I _{(A)}/2 π*r_{(m)}**

Where,

B→ magnetization of the magnitude A/m

μ_{0} →Permeability of free space 4 μ*10^{-7} T m/A

I→ Magnitude of the electric current in A

R→ distance in m

## Sample Sums :

### Sums 1:

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

### Solution :

In order to Find the magnetic field strength inside the solenoid, one must use B= μ_{0 } N_{i} .Furthermore one must note the number of loops per unit length

n=N/r=2000/2=1000m^{-1}=10cm^{-1}

Now one must substitute known values

B= μ_{0 } N_{i}

=(4 π 10^{-7} m/A) (1000m^{-1})(1600A)

=2.01T

### Sums 2:

Calculate the magnetic field strength inside a solenoid which is 3 m long and has 3000 loops. Furthermore, it carries a 1200 A current?

### Solution :

In order to Find the magnetic field strength inside the solenoid, one must use B= μ_{0 } N_{i} .Furthermore one must note the number of loops per unit length

n=N/r=3000/3=1000m^{-1}=10cm^{-1}

Now one must substitute known values

B= μ_{0 } N_{i}

=(4 π 10^{-7} m/A) (1000m^{-1})(1200A)

=1.501T