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

The correct option is B $\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$


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 (part 1 and 2) $B_1=B_2=\frac{{\mu }_0}{4\pi }.\frac{2i_{1}sin{45}^{\circ }}{a/2}\otimes =\frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}{i}_1}{a}\otimes$
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) $B_3=B_4=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{2}{I}_2}{a}\odot$
Also $i_1=2i_2$. So $(B_1=B_2)>(B_3=B_4)$
Hence net magnetic field at centre O
$B_{net}=(B_1+B_2)-(B_3+B_4)$
$=2\times \frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}\times \left(\frac{2}{3}i\right)}{a}-\frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}\left(\frac{i}{3}\right)\times 2}{a}$
$=\frac{{\mu }_0}{4\pi }.\frac{4\sqrt{2}i}{3a}(2-1)\otimes =\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$