Question

The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.

Answer 1

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  $B \alpha \frac{1}{r^2}$

The magnetic field is a volume of vectors

And $\phi\ bds = 0$. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the $B \alpha \frac{1}{r^2}$ it is only for the simple current-carrying wire for both finite or infinite length.