Ampere’s Law Formula

Andre-Marie Ampere gave the Ampere’s law in 1826. This is one of the basic law of magnetism which talks about the sum magnetic field through a closed current carrying hoop. Ampere’s law can be valuable when calculating magnetic fields of current distributions with a high degree of symmetry.

The magnetic field in space an electric current is proportionate to the electric current which works as its source, just as the electric field in space is proportional to the charge which serves as its source. Ampere’s Law says that for any closed loop path, the total of the length elements times the magnetic field in the direction of the length element is equivalent to the permeability times the electric current bound in the loop.  The law can be articulated in two arrangements, the “integral form” and the “differential form.”

The Ampere’s law formula is,

$\oint\vec{B}.\vec{dl}=\mu&space;_{0}I$

Where the permeability of the medium is mu and the magnetic field is ,

the infinitesimal length is , the current flowing through the closed loop is I .

Solved Samples – Ampere’s Law

Now we’ll discuss the problems connected to the Ampere’s law:

Question 1: Compute 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 parameters are,
R = 0.05m and I = 2amp and we know that, μ0μ0 = 4π×10−7π×10−7N/A2
The Ampere’s law formula is,

$\oint\vec{B}.\vec{dl}=\mu&space;_{0}I$

In the case of long straight wire

$\oint&space;\vec{dl}=2\Pi&space;R=2\times3.14\times&space;0.05=0.314$

$B\oint\vec{dl}=\mu&space;_{0}I$

$\vec{B}=\frac{\mu&space;_{0}I}{2\pi\,R}$

$\vec{B}=\frac{4\pi\times&space;10^{-7}\times&space;2}{0.314}=8\times&space;10^{-6}\,&space;T$

Question 2: Compute the magnetic field of a straight wire which is made as a circular loop with a radius of 0.1m. The current flowing through this secure loop is given as 5amp?

Solution:

Known parameters are,
R = 0.1m and I = 5amp and it’s given that,

μ0 = 4π×10−7π×10−7N/A2
The Ampere’s law formula is,
$\oint\vec{B}.\vec{dl}=\mu&space;_{0}I$

In the case of long straight wire

$\oint&space;\vec{dl}=2\pi&space;R=2\times&space;3.14\times&space;0.1=0.628$

So,$B\oint\vec{dl}=\mu&space;_{0}I$

$\vec{B}=\frac{\mu_{0}&space;I}{2\pi&space;R}$

$\vec{B}=\frac{4\pi\times&space;10^{-7}\times&space;5}{0.628}=1\times&space;10^{-5}\,&space;T$