Question

Find magnetic field at the point $P$.

• A. $\frac{(\sqrt{2}\pi +2){\mu }_{0}I}{8\sqrt{2}\pi R}$
• B. $\frac{3(\sqrt{2}\pi +1){\mu }_{0}I}{8\sqrt{2}\pi R}$
• C. $\frac{(3\sqrt{2}\pi +2){\mu }_{0}I}{8\sqrt{2}\pi R}$
• D. $\frac{(\sqrt{2}\pi +1){\mu }_{0}I}{4\sqrt{2}\pi R}$

Answer 1

The correct option is C $\frac{(3\sqrt{2}\pi +2){\mu }_{0}I}{8\sqrt{2}\pi R}$

Magnetic field $B$ at point $P$ due to circular path

$B_{arc}=\frac{{\mu }_{0}I}{2R}\times \frac{3}{4}\text{Inwards}(⨂)\left[\text{Fleming's right-hand thumb rule}\right]$

Since the point $P$ lies on the extended lines $AB$ and $CQ$ , the magnetic field at point $P$ due to $QC$ & $AB$ will be zero.

Magnetic field due to $AD$ at point $P$ be,


$B_{AD}=\frac{{\mu }_{0}I}{4\pi \left(2R\right)}\times (\sin {45}^{\circ }+\sin 0)$

$\Rightarrow B_{AD}=\frac{{\mu }_{0}I}{8\sqrt{2}\pi R}$$⨂$

Magnetic field due to $QD=$ Magnetic field due to $AD$

$\therefore B_{QD}=\frac{{\mu }_{0}I}{8\sqrt{2}\pi R}$$⨂$

Now, the net magnetic field at point $P$

$B_P=B_{arc}+B_{AD}+B_{DQ}$

$\Rightarrow B_P=\frac{3{\mu }_{0}I}{8R}+\frac{{\mu }_{0}I}{8\sqrt{2}\pi R}+\frac{{\mu }_{0}I}{8\sqrt{2}\pi R}$

$\Rightarrow B_P=\frac{(3\sqrt{2}\pi +2){\mu }_{0}I}{8\sqrt{2}\pi R}(⨂)$

Hence, option (c) is the correct answer.