Magnetic scalar potential is defined as
$U ({\vec{r}}_{2}) - U ({\vec{r}}_{1}) = - \int_{{\vec{r}}_{1}}^{{\vec{r}}_{2}} \vec{B} . \vec{d l}$
Apply this equation to a closed curve enclosing a long straight wire. The RHS of the above equation is then −μ₀ i by Ampere's law. We see that $U ({\vec{r}}_{2}) \neq U ({\vec{r}}_{1})$ even when ${\vec{r}}_{2} = {\vec{r}}_{1}$ . Can we have a magnetic scalar potential in this case?

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Solution

No, we cannot have a magnetic scalar potential here.

Ampere's law is a method of calculating magnetic field due to current distribution. On the other hand, magnetic scalar potential requires a magnetic field due to pole strength m.
Potential at a distance r is given by
$\frac{μ_{0} m}{4 π r}$
As there is no current distribution, no magnetic field due to poles or the pole strength is present. That is why we cannot have a magnetic scalar potential in this case.

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∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{u_{1}} & \frac{1}{u_{2}} & \frac{1}{u_{3}} a_{2} a_{3} - b_{2} b_{3} & a_{3} a_{1} - b_{3} b_{1} & a_{1} a_{2} - b_{1} b_{2} a_{2} b_{3} + a_{3} b_{2} & a_{3} b_{1} + a_{1} b_{3} & a_{1} b_{2} + a_{2} b_{1} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

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(x + a)^{n} = x^{n} +^{n} C_{1} x^{n - 1} a +^{n} C_{2} x^{n - 2} a^{2} + . . . + a^{n}

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