Suppose that there is a semi-infinite wire which extends to infinity only in one direction. There are no other circuit elements at the other end(finite end) of the wire and the current does not loop. The magnetic field obviously has cylindrical symmetry when the Amperian contour is taken as a circle with its center on the wire.
However, due to charge accumulation there is a time-dependent electric field; hence a displacement current. How can I formulate the Ampere's law and show that the magnetic field is the half of that the infinite wire at the finite end of the wire? Do you think treating charge accumulation as a point charge with changing amount of charge right at the finite end of the wire will suffice?
The original question is: 
Answer
Ampere's law (for a steady current) states that $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I$$
If we consider an infinite wire, then symmetry tells us that the B-field at the point $A$ and all other points on a circle of radius $(R+y)$ is constant in magnitude and is in the azimuthal direction. Hence the magnitude of the B-field is given by $$ 2 \pi (R+y)B = \mu_0 I$$ $$ B = \frac{\mu_0 I}{2\pi(R+y)}$$
So now, for a semi-infinite wire, I take away half the wire and hence half the vector field. But, before I just say that the new field is half of the original one, I need to establish that the B-fields from each "half" of the infinite wire are in fact in the same direction so that they add in parallel fashion. The Biot-Savart law tells us that each wire element produces a B-field that is perpendicular to the current and perpendicular to a displacement joining the wire element and the point at which I wish to know the field. So, the B-field is always in a direction azimuthal to the wire, whichever piece(s) of wire we consider. This means that the new B-field for the semi-infinite wire is in the same direction as for the full infinite wire but has half the magnitude.
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