Verify the Ampere's law for magnetic field of a point dipole of dipole moment `overset(to)(M) = M hat(k)`. Take C as the closed curve running clockwise along
(i) the z-axis from `z=a gt 0` to` z= R,`
(ii) along the quarter circle of radius R and centre at the origin, in the first quadrant of x-z plane,
(iii) along the x-axis from x = R to x = a and
(iv) along the quarter circle of radius a and centre at the origin in the first quadrant of x-z plane.

Explanation

From P to Q, every point on the z-axis lies at the axial line of magnetic dipole of moment `overset(to)(M)`. Due to this magnetic moment the point (0, 0, Z) at z distance the magnetic field induction,
`B = 2 ((mu_0)/( 4 pi) (M)/( z^3))`
`B= (mu_0 M)/(2 pi z^3)`

(i) From Ampere.s law,
At point from P to Q along z-axis
`int_(P)^(Q) overset(to)(B). overset(to)(d) l = int_(P)^(Q) B dl cos 0^(@) = int_(a)^(R) B dz`
`= int_(a)^(R) (mu_0)/( 2pi ) (M)/(z^3) dz = (mu_0 M)/( 2pi) (-(1)/(2) ) ((1)/(R^2) - (1)/(a^2) )`
`= (mu_0 M)/( 4 pi ) ((1)/( a^2) - (1)/(R^2))`
(ii) Along the quarter circle QS of radius R is given in the figure below.

The point A lies on the equatorial line of the magnetic dipole of moment M `sin theta`. Magnetic field at point A on the circular arc is,
`B =(mu_0)/( 4pi ) (M sin theta)/( R^3) dl and dl = R d theta`
`therefore d theta = (dl)/( R) `
From Ampere.s law,
`therefore int overset(to)(B). overset(to)(dl)= int B dl cos theta`
`= int_(0)^(pi //2) ((mu_0)/(4pi ) (M sin theta)/(R^3)) R d theta`
`= (mu_(0) M)/( 4 pi R^2) int_(0)^(pi //2)sin theta d theta`
`= (mu_0)/( 4pi ) (M)/( R) [ - cos theta ]_(0)^(pi//2)`
`= (mu_0)/( 4pi ) (M)/(R ) [- cos (pi)/(2) + cos 0^(@) ]`
`= (mu_0 M)/( 4pi R^2 ) = [0+1]`
`= (mu_0 M)/( 4pi R^2)`
(iii) From Ampere.s law, along x-axis over the path ST, consider the figure given ahead.

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance x from the dipole is,
`B= (mu_0)/( 4pi ) (M)/( x^3)`
`therefore int_(S)^(T) overset(to)(B). overset(to)(d) l = int_(R)^(a) - (mu_0 overset(to)(M) )/( 4pi x^3) . overset(to)(d) l = 0" "(because "Angle between" overset(to)(M) and overset(to)(d) l "is" 90^@)`
`int_(S)^(T) overset(to)(B). overset(to)(d) l = (- mu_0)/( 4pi x^3) int_(R)^(a) dl cos 90^@`
(iv) Along the quarter circle TP of radius a. Consider the figure given below,

From case II line integral of `overset(to)(B)` along the quarter circle TP of radius a is circular arc TP.
`int overset(to)(B). overset(to)(d) l = int_(pi//2)^(0) (mu_0)/( 4pi ) (M sin theta )/( a^3) ad theta`
`= (mu_0)/( 4pi ) (M)/(a^2) int_(pi//2)^(0) sin theta d theta = (mu_0 M)/( 4pi a^2) [ - cos theta ]_(pi//2)^(0)`
`=(mu_0 M)/(4pi a^2) [- cos 0^@ + cos (pi)/(2) ]`
`= (mu_0 M)/( 4pi a^2) [-1 + 0]`
`= - (mu_0)/(4pi) (M)/(a^2)`
`therefore` For whole closed path PQST,
`therefore oint_("PQST") overset(to)(B).overset(to)(d) l = int_(P)^(Q) overset(to)(B).overset(to)(d) l + int_(Q)^(S) overset(to)(B). overset(to)(d) l + int_(T)^(P) overset(to)(B). overset(to)(d)l`
`(mu_0 M)/(4pi) [ (1)/(a^2) - (1)/(R^2) ] + (mu_0)/(4pi ) (M)/(R^2) + 0 + (-(mu_0)/( 4pi ) (M)/( a^2) )`
`therefore oint_("PQST") overset(to)(B). overset(to)(d)l = (mu_0 M)/( 4pi a^2) - (mu_0 M)/( 4pi R^2) + (mu_0 M)/( 4pi R^2) - (mu_0 M)/( 4pi a^2) `
`=0`