## Ampere’s Law Formula :

Ampere’s Law formula B_{(}_{N/m)} in Newton per meter and multiply the dl is equal to the μ_{0(}_{N/A2)} in Newton per Ampere and multiply the current flowing through the closed loop I_{(amp)} in ampere. Hence the Ampere’s Law formula can be written as,

**∮B _{(}_{N/m)} *dl=μ_{0(}_{N/A2)} * I_{(amp)}**

Natation’s used in Ampere’s Law Formula

B →magnetic field in N/m

L → infinitesimal length in m

I → current flowing through the closed-loop in amp

μ→ e permeability in N/A^{2}

### Example 1:

Calculate the magnetic field of a long straight wire that has a circular loop with a radius of 0.08m. 4amp is the reading of the current flowing through this closed loop.

### Solution :

Given,

R=0.08m

I=4amp

μ_{0} = 4π×10^{-7}N/A^{2}

Ampere’s law formula is

∮B dl=μ_{0} I

In the case of long straight wire

∮ dl =2π R

=2 ×3.14×0.08

=0.5024

B∮ dl= μ_{0} I

B= μ_{0} I /2π R

B=4 π ×10^{-7} ×4 /0.5024

=0.507×10^{-7} T

### Example 2:

Calculate the magnetic field of a long straight wire that has a circular loop with a radius of 0.05m. 2amp is the reading of the current flowing through this closed loop.

### Solution :

Given,

R=0.05m

I=2amp

μ_{0} = 4π×10^{-7}N/A^{2}

Ampere’s law formula is

∮B dl=μ_{0} I

In the case of long straight wire

∮ dl =2π R

=2 ×3.14×0.05

=0.314

B∮ dl= μ_{0} I

B= μ_{0} I /2π R

B=4 π ×10^{-7} ×2 /0.314

=8×10^{-6} T