Ampere's Law
Magnetics of Class 12
The integral of the quantity . daround loop is related to the current flowing through the surface bounded by the loop.
∮. d = μoI (3.22)
Note that there are often contribution to from currents not enclosed by the loop. This law may be used to find 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) ∮. d = ∫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 = μ_{0}I/2πr 

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.
∮. d = μoI
∫_{a}^{b}. d +∫_{d}^{c} . d + ∫_{c}^{d}. d + ∫_{d}^{a}. d = μoI
Since ⊥ d in the path b → c and d → a, therefore
∫_{b}^{c}. d = ∫_{d}^{a}. d= 0
Moreover along the paths a → b and c → d, is parallel to path directions. Thus
∫_{a}^{b}. d + ∫_{c}^{d}. d = 2Bl
The current enclosed is I = j l
Thus, applying Ampere’s law
2B l = μo(j l)
or B = μ_{0}j/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 = μ_{0}NI/2πr a ≤ r ≤ b (3.24)
Note that the magnetic field varies as I / r within a toroid.