Question: Figure 29-28 shows three circuits consisting of straight radial lengths and concentric circular arcs (either half- or quarter-circles of radii r, 2r, and 3r). The circuits carry the same current. Rank them according to the magnitude of the magnetic field produced at the center of curvature (the dot), greatest first.

• Figure of three circuits is given.
• Same current is flowing through each circuit.

Formulae are as follows:

Find the net magnetic field at the center of each circuit using the formula for the magnetic field at the center of a circular loop. Comparing them rank the givencircuits according to the magnitude of the magnetic field at the center.

Where, B is a magnetic field, R is the radius, iis current,𝛍is permeability.

The magnitude of magnetic field at the center of arc is given by,

$B=\frac{{\mu }_{0}i}{4\pi R}$

Let the magnetic field at the center of arc due to arc of radius r be B₁ and due to arc of radius 2r be B₂,and due to arc of radius 3r be B₃ .

Then,

The net magnetic field at the center of circuit (a) is,

$B=B_1+B_3=\frac{{\mu }_{0}i\pi }{4\pi r}+\frac{{\mu }_{0}i\pi }{12\pi r}=\frac{{\mu }_{0}i}{3r}$

The net magnetic field at the center of circuit (b) is,

$B=B_1-B_3=\frac{{\mu }_{0}i\pi }{4\pi r}-\frac{{\mu }_{0}i\pi }{12\pi r}=\frac{{\mu }_{0}i}{6r}$

The net magnetic field at the center of circuit (c) is,

$B=B_1+B_2+B_3=\frac{{\mu }_{0}i\pi }{8\pi r}+\frac{{\mu }_{0}i\pi }{16\pi r}+\frac{{\mu }_{0}i\pi }{12\pi r}=\frac{13{\mu }_{0}i}{48r}$

Hence, the ranking of circuits according to the magnitude of the net magnetic field at the center is,$a>c>B$

Therefore, the magnetic field at the center of arc depends inversely on its radius.