A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of 10 A. The magnetic field at point O will be $\mathrm{a} \mathrm{x} {10}^{-\mathrm{b}}\mathrm{T}$. Then the value of a+b is(where a,b are integers)

From the given figure as shown below

The magnetic field at point O due to wires PQ and RS will be zero.

Magnetic field due to arc QR at point O will be

          ${\mathrm{B}}_1=\frac{\mathrm{\theta }}{2\mathrm{\pi }}\left(\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{a}}\right)$

Here, $\mathrm{\theta }={45}^{\circ }=\frac{\mathrm{\pi }}{4}\mathrm{rad},\mathrm{i}=10\mathrm{A}$

and $\mathrm{a}=3\mathrm{cm}=3\times {10}^{-2}\mathrm{m}$

$\begin{array}{c}\Rightarrow {\mathrm{B}}_1=\frac{\mathrm{\pi }}{2\mathrm{\pi }\times 4}\left(\frac{{\mathrm{\mu }}_0\times 10}{2\times 3\times {10}^{-2}}\right)\\ =\frac{{\mathrm{\mu }}_0\times 5}{2\times 12\times {10}^{-2}}=\frac{5\times {\mathrm{\mu }}_0\times {10}^2}{24}\end{array}$

Direction of field B₁ will be coming out of the plane of figure.

Similarly, field at point O due to arc SP will be

$\begin{array}{c}{\mathrm{B}}_2=\frac{\mathrm{\pi }}{4}\left(\frac{1}{2\mathrm{\pi }}\right)\left[\frac{{\mathrm{\mu }}_0\times 10}{2\times (2+3)\times {10}^{-2}}\right]\\ =\frac{{\mathrm{\mu }}_0\times {10}^2}{8}\end{array}$

Direction of B₂ is going into the plane of the figure.

$\therefore$The resultant field at O is

$\begin{array}{l}\mathrm{B}={\mathrm{B}}_1-{\mathrm{B}}_2=\frac{1}{2}\left(\frac{5\times {\mathrm{\mu }}_0}{12\times {10}^{-2}}-\frac{{\mathrm{\mu }}_0}{4\times {10}^{-2}}\right)\\ =\frac{4\mathrm{\pi }\times {10}^{-7}}{12\times {10}^{-2}}\cong 1\times {10}^{-5}\mathrm{T}\end{array}$