Question

Figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of magnetic field at the centre of the loop assuming uniform wire is

• A. $\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\odot$
• B. $\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$
• C. $\frac{\sqrt{2}{\mu }_{0}i}{\pi a}\odot$
• D. $\frac{\sqrt{2}{\mu }_{0}i}{\pi a}\otimes$

Answer 1

Answer: (B)

According to question resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flows through $ADC$ is half that of $ABC$
i.e. $\frac{i_2}{i_1}=\frac{1}{2}$.
Also $i_1+i_2=i$
$\Rightarrow i_1=\frac{2i}{3}$ and $i_2=\frac{i}{3}$
Magnetic field at centre $O$ due to wire $AB$ and $BC$ (parts 1 and 2) $B_1=B_2=\frac{{\mu }_0}{4\pi }\cdot \frac{2i_1\sin 45^{\circ }}{a/2}\otimes =\frac{{\mu }_0}{4\pi }\cdot \frac{2\sqrt{2}{i}_1}{a}\otimes$ and magnetic field at centre $O$ due to wires $AD$ and $DC$ (i.e. parts 3 and 4) $B_3=B_4=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{2}{i}_2}{a}$
Also $i_1=2i_2$. So $\left(B_1=B_2\right)>\left(B_3=B_4\right)$
Hence net magnetic field at centre $O$
$B_{net}=\left(B_1+B_2\right)-\left(B_3+B_4\right)$
$=2\times \frac{{\mu }_0}{4\pi }\cdot \frac{2\sqrt{2}\times \left(\frac{2}{3}i\right)}{a}-\frac{{\mu }_0}{4\pi }\cdot \frac{2\sqrt{2}\left(\frac{i}{3}\right)\times 2}{a}$
$=\frac{{\mu }_0}{4\pi }\cdot \frac{4\sqrt{2}i}{3a}(2-1)\otimes =\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$