A current is set up in a wire loop consisting of a semicircle of radius$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$,a smaller concentric semicircle, and tworadial straight lengths, all in the same plane. Figure 29-47ashows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is $\mathbf{47}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{\mu T}$. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (Figure29-47 b).The magnetic field produced at the (same) center of curvature now has magnitude $\mathbf{15}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{\mu T}$, and its direction is reversed. What is the radius of the smaller semicircle.

Given the magnetic field at the center of the circuit in both cases, find the magnetic field due to a large circle and magnetic field due to a small circle. Find the relation between the magnetic field and radius of the circle. From this, and find the radius of the smaller circle.

Formula:

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

Let the magnetic fields produced at the center of circuit due to larger and smaller circle be $B_L$and $B_S$respectively.

From figure a, we can say that,

Magnetic field produced at the center of the circuit will be the sum of the magnetic fields produced by the larger and smaller circle because the direction of the current flowing through both arcs is same. So,

$B_a=B_L+B_S$

$B_L+B_S=47.25\mu T$ ……. (1)

From figure b, we can say that,

The magnetic field produced at the center of the circuit will be the difference between the magnetic fields produced by larger and smaller circle, because the direction of current flowing through both arcs is opposite. So,

$B_b=B_L-B_S$

$B_S-B_L=15.75\mu T$ ……. (2)

Solve equation 1 and 2 simultaneously.

$B_L=15.75\mu T$and $B_S=31.5\mu T$

Consider the equation of the magnetic field as:

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

The current flowing through both the circles is same.

So $\frac{{\mu }_{0}I\varphi }{4\pi }$will be constant for both circles.

Consider the relation is obtained as:

$B\propto \frac{1}{R}$

Solve further as:

$B_L{R}_L=B_S{R}_S$

$R_S=\frac{B_L{R}_L}{B_S}$

From the equation (1) and (2) substitute the values as:

$R_S=\frac{15.75\times 0.04}{31.5}$

$$\begin{gathered} R_S=0.02\text{m} \\ {R}_S=2\text{cm} \end{gathered}$$

Thus, the radius of the smaller semicircle is$2\mathrm{cm}$.