Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral `(L)=|int_(-L)^(L)vecB*vec(dl)|` taken along z-axis.
(a) Show that (L) monotonically increases with L.
(b) Use an appropriate Amperian loop to show that `(oo)=mu_(0)I`, where I is the current in the wire.
( c) Verify directly the above result.
(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about (L) and `(oo)`?

Explanation

(a) B(z) is same at every point on z-axis. So, is monotonically function for L.
`vecBandvec(dl)` are in same direction. So,
`vecB*vec(dl)=Bdlcos0=Bdl`
(b) (L) + by circumference at large distance
`C=mu_(0)I`
Now, `Ltooo`
At large distance `to0" "(becauseBprop1/r^(3))`
`(oo)-mu_(0)(I)`
( c)
Current carrying ring of radius R is in xy-plane. Centre of it is at origin point, magnetic field at any point from centre of loop is,
`B_(z)=(mu_(0)IR^(2))/(2(z^(2)+R^(2))^(3/2))`
`int_(-oo)^(oo)B_(z)dx=int_(-oo)^(oo)(mu_(0)IR^(2))/(2(z^(2)+R^(2))^(3/2))dz`
Taking `z=Rtantheta`
`dz=Rsec^(2)thetad theta`
`int_(-oo)^(oo)B_(z)dz=(mu_(0)I)/2int_((-pi)/2)^((+pi)/2)(costheta)d theta=mu_(0)I`
(d) `B(z)_("square")ltB(z)_("circular loop")`
`thereforeJ(L)_("square")ltJ(L)_("circular loop")`
but as per discussion of option (b),
`J(oo)_("square")=J(oo)_("circular loop")`