Magnetic field due to current carrying wire at the wire [closed]

What is the magnetic field due to current carrying wire at a point on the wire itself?

closed as unclear what you're asking by Chris♦, John Duffield, Jon Custer, sammy gerbil, M. EnnsMar 21 '18 at 20:20

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• -1 Not clear what your difficulty is. – sammy gerbil Mar 19 '18 at 12:19

Complementing jim's answer in order to explicitly address the questions that possibly motivated the original post:

• does the field intensity diverge at the wire?
• where does it point to?

The answer is that in practice we have a current density ($I(A)/A$); and, as long as this density is finite, considering a position $r\to 0$ necessarily leads to the magnetic field at this position being that generated by a vanishing amount of current and, thus, having vanishing intensity: $\mathbf{B}\to 0$.

That is, instead of diverging, the magnetic field is zero and, in particular, has no defined direction.

What about $B =\frac{\mu_0 I}{2\pi r}$, then? Well, if you try to apply it for $r\to 0$, you approach the limit of the wire being a mathematical line, with zero radius and area, and that implies, if the current is finite, an infinite current density (finite $I$ going through a vanishing cross-section area). So you're assuming a diverging physical situation, it's then not surprising you get other divergences.

• From given eqn B is inversely proportional to r. If we keep current constant and plot a graph of B vs r we get a graph similar to y=1/x for which y and limit as x=0 or x->0 is not defined. Is this what you mean to say. Thanks – Dhruv Deshmukh Mar 19 '18 at 10:55
• @DhruvDeshmukh, Yes, the limit $x\to 0$ of the expression for $\mathbf{B}$ is divergent. – stafusa Mar 19 '18 at 11:24

For an infinitely long wire of infinitesimal thickness carrying a steady current $I$ the magnetic field at a distance $r$ from the wire is $$B =\frac{\mu_0 I}{2\pi r}.$$ This result can be derived from Ampere's Law $$\int {\bf B . d s} = \mu_0 \times \text{current enclosed by path},$$ where the current enclosed by the path (for infinitesimal thickness the enclosed current is always the total current $I$ flowing through the wire. For a wire of finite thickness you can still make use of Ampere's Law though for a distance $r \lt s$ the enclosed current is now only a fraction of the total current flowing through the wire. Typically this is taken to be $$I \frac{r^2}{s^2}.$$ You can then determine the magnetic field for the two cases (i) $r \le s$ (current = $I \frac{r^2}{s^2}$) and (ii) $r \ge s$ (current = $I$).

For distances inside the wire you only have a fraction of the current that contributes to the magnetic field and the magnetic field has a finite value at a point on the wire itself.

• What I want to know is if wire has no thickness i.e. a 1d straight line which we normally use for problem solving then what is the B at a point on the wire. – Dhruv Deshmukh Mar 19 '18 at 10:53
• But this is not physical, real wires have thickness. – jim Mar 19 '18 at 19:44