Suppose that there is a long wire carrying a current. According to Ampere's law, there one can determine the magnetic field (B) on a circular path around the wire given its radius (R).

Consider the locations on the bold line in the figure, the total magnetic "effect" (as in Ampere's law) should be B times the length of the bold line, which should be a finite number.

enter image description here

If I bend the wire into half a circle of radius R, then all those points on the bold line should converge into a single point C, but in that case wouldn't its B become infinity as it has zero length but a finite magnetic "effect"?

  • $\begingroup$ In Ampere's law you have an integral over a closed path. Where is this closed path in your example? $\endgroup$
    – nasu
    Feb 13, 2023 at 4:27
  • $\begingroup$ the closed path is just the individual circles, then by the principle of superimposition I add the magnetic "effect" together. Is this incorrect? $\endgroup$
    – Kent Tong
    Feb 13, 2023 at 4:36
  • $\begingroup$ The bold line is not a closed path. What is the relation between it and Ampere's law? "Magnetic effect" is not a standard quantity. How do you define it? $\endgroup$
    – nasu
    Feb 13, 2023 at 4:52
  • $\begingroup$ The magnetic field at the center of a circular loop or just for an arc-shaped wire is finite. $\endgroup$
    – nasu
    Feb 13, 2023 at 4:54
  • $\begingroup$ See : hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html $\endgroup$
    – The Tiler
    Feb 13, 2023 at 13:07

1 Answer 1


(a) "the total magnetic "effect" (as in Ampere's law) should be B times the length of the bold line" No: the Ampère's law line integral is of the field component along the chosen line, but in your case, $\mathbf B$ is at right angles to the bold line. Setting this aside, I'm afraid that your argument, starting with field strengths at different points along a line, doesn't have a sound basis and isn't valid. [See, for example, nasu's comment, "magnetic effect is not a standard quantity. How do you define it?"] That's why the prediction of an infinite field at the centre of the semicircle is wrong.

(b) For the straight wire the field at a point (say on your bold line) is not due just to the element of wire nearest that point, but due to the whole of the wire (though points further away contribute less, in accordance with the Biot-Savart law). The centre of an arc is not the only place where fields from different elements of a wire re-inforce.

(c) Bending the wire into a semicircle (of radius $R$) will increase the field somewhat at the centre, O, of the semicircle, compared with its value at the same distance, $R$, from the centre of a straight wire of the same length ($\pi R$). This is because no element of the wire will be further than $R$ from O, and because $\sin\theta$ in the B–S law will be 1 for all elements of the semicircle. It is a good exercise to evaluate the field in the two cases, using the Biot–Savart law. [I find that for the semicircle, $B=\frac{\mu_0I}{4R}$, whereas for the straight wire $B=\frac{\mu_0I}{7.45\ R}$.]


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.