91Ó°ÊÓ

The radius of the bigger semicircle =$R=10.0cm$

The net magnetic field at the center in case a =$B_b=12.0\mathrm{μ}\text{T}$

The net magnetic field at the center in case b = $B_a=10.0\mathrm{μ}\text{T}$

The magnetic field at a point due to a current-carrying wire is determined using the Biot-Savart law. The direction of the magnetic field is decided by the right-hand rule. The net magnetic field at the point is the vector sum of the magnetic fields due to all the wires.

Formula:

$B=\frac{{\mathrm{μ}}_{0}i\mathrm{ϕ}}{4\mathrm{π}R}$

The radial lengths of the loop do not produce any magnetic field at the center of the curvature. Hence the net magnetic field at the center of the curvature is the vector sum of the magnetic fields of the two semicircles.

In case a, both the semicircles are in the same plane. Thus, the direction of the current will be opposite in the two semicircles. Hence according to the right hand rule, the direction of the magnetic field at the center due to both semicircles will be the same. Hence the net magnetic field will be addition of the two fields

For semicircular loop, $\mathrm{θ}=\mathrm{π}$

For case b, the two semicircles are in planes perpendicular to each other. The direction of the current will be opposite in the two semicircles. Hence according to the right hand rule, the direction of the magnetic field at the center due to both semicircles will be the same (i.e., out of page or into the page) but perpendicular to each other. Hence the net magnetic field will be the vector sum of the two fields.

Then the magnitude of the net magnetic field will be calculated as

$$\begin{gathered} B_b^2=B_B^2+B_s^2 \\ {B}_b^2={\left(\frac{{\mathrm{μ}}_{0}i\mathrm{π}}{4\mathrm{π}R}\right)}^2+{\left(\frac{{\mathrm{μ}}_{0}i\mathrm{π}}{4\mathrm{π}r}\right)}^2 \\ {B}_b^2={\left(\frac{{\mathrm{μ}}_{0}i}{4R}\right)}^2\left(\frac{1}{R^2}+\frac{1}{r^2}\right) \\ {B}_b=\left(\frac{{\mathrm{μ}}_{0}i}{4R}\right)\sqrt{\left(\frac{1}{R^2}+\frac{1}{r^2}\right)}······\left(2\right) \\ =10.0×{10}^{-6} \end{gathered}$$

Taking ratio of equations (1) and (2), we get

$$\begin{gathered} \frac{B_a}{B_b}=\frac{12}{10}=\frac{\frac{{\mathrm{μ}}_{0}i}{4}\left(\frac{1}{R}+\frac{1}{r}\right)}{\left(\frac{{\mathrm{μ}}_{0}i}{4R}\right)\sqrt{\left(\frac{1}{R^2}+\frac{1}{r^2}\right)}} \\ 1.2=\frac{\frac{R+r}{rR}}{\sqrt{\left(\frac{R^2+r^2}{R^2{r}^2}\right)}}=\frac{R+r}{rR}×\frac{rR}{\sqrt{\left(R^2+r^2\right)}} \\ 1.2=\frac{R+r}{\sqrt{\left(R^2+r^2\right)}} \\ \left(R+r\right)=1.2\sqrt{\left(R^2+r^2\right)} \end{gathered}$$

$$\begin{gathered} {\left(R+r\right)}^2=1.44\left(R^2+r^2\right) \\ \left(R^2+r^2+2Rr\right)=1.44\left(R^2+r^2\right) \\ \left(2Rr\right)=1.44\left(R^2+r^2\right)-\left(R^2+r^2\right) \\ 2Rr=0.44\left(R^2+r^2\right) \end{gathered}$$

Now we put the value of R in this equation and reduce it to

$$\begin{gathered} 2×10r=0.44\left({10}^2+r^2\right) \\ 0.44r^2-20r+44=0 \end{gathered}$$

We solve this quadratic equation using the standard formula:

$x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

Here, a = 0.44, b = 20 and c = 44

$$\begin{gathered} r=\frac{20Â\pm \sqrt{{20}^2-\left(4×0.44×44\right)}}{2×0.44} \\ r=\frac{20Â\pm \sqrt{\left(400\right)-\left(77.44\right)}}{0.88v} \\ r=\frac{20Â\pm \sqrt{322.56}}{0.88}=\frac{20Â\pm 17.96}{0.88} \\ r=\frac{2.04}{0.88}or\frac{37.96}{0.88} \\ r=2.32\text{cm}or43.1\text{cm} \end{gathered}$$

But the radius of this semicircle must be smaller than 10.0, so r = 43.1 cm is not an acceptable solution.

Hence r = 2.32 cm