Question

Figure shows a square loop $A B C D$ with edge length $a$. The resistance of the wire $A B C$ is $r$ and that of $A D C$ is $2 r$. 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 $A D C$ is twice that of wire $A B C$. Hence current flows through $A D C$ is half that of $A B C$
i.e. $\frac{i_{2}}{i_{1}}=\frac{1}{2}$.
Also $i_{1}+i_{2}=i $
$\Rightarrow i_{1}=\frac{2 i}{3}$ and $i_{2}=\frac{i}{3}$
Magnetic field at centre $O$ due to wire $A B$ and $B C$ (parts 1 and 2) $B_{1}=B_{2}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 i_{1} \sin 45^{\circ}}{a / 2} \otimes=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \sqrt{2} i_{1}}{a} \otimes$ and magnetic field at centre $O$ due to wires $A D$ and $D C$ (i.e. parts 3 and 4) $B_{3}=B_{4}=\frac{\mu_{0}}{4 \pi} \frac{2 \sqrt{2} i_{2}}{a}$
Also $i_{1}=2 i_{2}$. 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} $
$=\frac{\mu_{0}}{4 \pi} \cdot \frac{4 \sqrt{2} i}{3 a}(2-1) \otimes=\frac{\sqrt{2} \mu_{0} i}{3 \pi a} \otimes$