4.7 Ampere's Circuital Law

Subject: Physics

Class XII

Chapter: 4. Moving Charges and Magnetism

Ampere's Law

 "The integral of magnetic field around a closed path is equal to the product of absolute permeability and total current passing through it."

Mathematical form

∮B.dl = μ₀.I

Here,

B = Magnetic field

dl = Current element

μ₀ = Absolute Permeability

I = Current passing through it

Proof

Consider about an infinitesimal carrying conductor through which I current passing is passing. Magnetic field lines are produced around the conductor as the form of concentric circles.

Magnetic field at a distance due to this infinitesimal conductor. 

B = μ₀.2I/4πa ................... (i)

Now consider about a circle of radius a having a current element dl.

Both B and dl have same direction because B is along the the tangent of circle.

Hence,

B.dl = Bdlcosθ = Bdlcos0⁰ = Bdl

Now integrate the closed path

∮B.dl = ∮Bdl .............. (ii)

Now putting the value of B

∮B.dl = ∮μ₀.2Idl/4πa

∮B.dl  = μ₀.2I/4πa∮dl

For complete cirecle

∮dl = 2πa

∮B.dl  = μ₀.2I.2πa/4πa

∮B.dl = μ₀.I

Proved

4.6 Magnetic Field on the Axis of Circular Current Loop

Subject: Physics

Class : XII

Chapter 4. Moving Charges and Magnetism

4.6 Magnetic Field on the Axis of Circular Current Loop

Consider about a circular loop of radius R carrying the current I. The loop is placed in the y-z plane with its centre at the origin O. The x axis is the axis of the loop. There is a point P at a distance x from the centre of the loop at which magnetic field is to be determined.

Mathematical Calculation

dl is the conducting element of the loop.

The magnitude of dB of the magnetic field due to dl is according to Biot-Savart Law

dB = 𝝁₀I|dlxr|/4𝜋r³
Now 
r² = x² + R²
Any element to loop will be perpendicular to the displacement vector r from dl to the axial point P is in the x-y plane.
Hence,
|dlxr| = rdl
∴dB = 𝝁₀Idlr/4𝜋r³ ............... (i)
Now putting the value of r
dB = 𝝁₀Idl/4𝜋(x² + R²)
The direction of dB is perpendicular to the plane formed by dl and r.
It has an x- component dBₓ and a component perpendicular to x-axis dB⫡.
When the components perpendicular to the x- axis are summed over, they cancel out.
Only the x component survives.
∴dBₓ = dBcos𝜃
From figure 
cos𝜃 = R/√(x² + R²) ............. (ii)
Now from eqn. no. (i) and (ii)
dBₓ = 𝝁₀IdlR/4𝜋(x² + R²)³/²
For whole circular loop
dl = 2𝜋R
Magnetic field at P due to entire circular loop
= B = Bₓi = 𝝁₀IR²/2𝜋(x² + R²)³/²i
Field at the centre of the loop. Here x = 0
B₀ = 𝝁₀I/2R

The direction of the magnetic field is determined by the help of 
Fleming Right Hand Thumb rule.