Ampere's Law

Magnetics of Class 12

The integral of the quantity Ampere's Law. dAmpere's Lawaround loop is related to the current flowing through the surface bounded by the loop.

Ampere's Law. dAmpere's Law = μoI (3.22)

Note that there are often contribution to Ampere's Law from currents not enclosed by the loop. This law may be used to find Ampere's Law if the geometry of the currents allow the choice of a path for which the integral may be easily evaluated.

Application of Ampere law

1. Infinitely long straight wire,

Consider a circular path of radius r as shown in the fig. (3.24)

Ampere's Law. dAmpere's Law = ∫Bdl cos0o = B ∫ dl = B (2πr)

Since the current enclosed by the circular path is I, therefore, using Ampere’s law

B (2πr) = μoI

or B = μ0I/2πr

Ampere's Law

2. Infinite current sheet

Consider an infinite large sheet of current as shown in the figure (3.25) where j is the linear current density i.e. the current per unit length. Imagine a rectangular loop around the sheet as shown.

Ampere's Law

Ampere's Law. dAmpere's Law = μoI

abAmpere's Law. dAmpere's Law +∫dc Ampere's Law. dAmpere's Law + ∫cdAmpere's Law. dAmpere's Law + ∫daAmpere's Law. dAmpere's Law = μoI

Since Ampere's Law⊥ dAmpere's Law in the path b → c and d → a, therefore

bcAmpere's Law. dAmpere's Law = ∫daAmpere's Law. dAmpere's Law= 0

Moreover along the paths a → b and c → d, Ampere's Law is parallel to path directions. Thus

abAmpere's Law. dAmpere's Law + ∫cdAmpere's Law. dAmpere's Law = 2Bl

The current enclosed is I = j l

Thus, applying Ampere’s law

2B l = μo(j l)

or B = μ0j/2 (3.23)

3. Toroidal Solenoid

A toroid consisting of N turns is shown in the figure (3.26 a).

The magnetic field lines are closed rings as shown in fig. (3.26 b).

Applying Ampere’s law to a loop of radius r

B ∫dl = B (2πr) = μo (NI)

Then B = μ0NI/2πr a ≤ r ≤ b (3.24)

Note that the magnetic field varies as I / r within a toroid.

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