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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. |
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Answer» Solution :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 sintheta )/( a^3) ad theta` `= (mu_0)/( 4pi ) (M)/(a^2) int_(pi//2)^(0) sin thetad 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` |
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