Physics - Magnetism Question with Solution | TestHub

Both straight wires lie on the horizontal line through A, so they contribute nothing at A. The vertical segment points radially at A, so it also contributes nothing.

The large arc is a quarter circle of radius R centred at A:$B_1=\frac{{\mu }_{0}i}{4R}\times \frac{90^{\circ }}{180^{\circ }}=\frac{{\mu }_{0}i}{8R}$

The small arc is a quarter circle of radius R/2 centred at A, carrying current in the same sense:$B_2=\frac{{\mu }_{0}i}{8(R/2)}=\frac{{\mu }_{0}i}{4R}$

Both fields point the same way at A, so they add:$B=\frac{{\mu }_{0}i}{8R}+\frac{{\mu }_{0}i}{4R}=\frac{3{\mu }_{0}i}{8R}$

With $i=\frac{20}{\pi }\mathrm{A}$ and R = 1 m:$B=\frac{3\times 4\pi \times 10^{-7}\times 20/\pi }{8\times 1}=3\times 10^{-6}\mathrm{T}=3\mu T$

Answer: 3