Question

Figure shows a square loop $\text{ABCD}$ with edge length $a$. The resistance of the wire $\text{ABC}$ is $r$ and that of $\text{ADC}$ is $2r$. The value of magnetic field at the center 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 the question resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flowing through $ADC$ is half that of $ABC$
$(\because i\propto \frac{1}{r})$

$\Rightarrow \frac{i_2}{i_1}=\frac{1}{2}$

Also $i_1+i_2=i$

or, $i_1+\frac{i_1}{2}=i$

$\therefore i_1=\frac{2i}{3}\text{and}{i}_2=\frac{i}{3}$


Let, the net magnetic field at center $\text{O}$ due to wires $\text{AB}$ and $\text{BC}$ be ${\text{B}}_1$

$\Rightarrow B_1=2\times \left[\frac{{\mu }_{0}i}{4\pi \left(d\right)}\right[sin{\varphi }_1+sin{\varphi }_2\left]\right]$

or, $B_1=\frac{2{\mu }_0{i}_1}{4\pi \left(a/2\right)}[sin{45}^{\circ }+sin{45}^{\circ }]$

or, $B_1=\frac{2{\mu }_0{i}_1}{2\pi a}\left[\frac{2}{\sqrt{2}}\right]=\frac{\sqrt{2}{\mu }_0{i}_1}{\pi a}\otimes$

Similarly the magnetic field at center $\text{O}$ due to wire $\text{AD}$ and $\text{DC}$ will be,

$B_2=2\times \left[\frac{{\mu }_0{i}_2}{4\pi d}\right(sin{45}^{\circ }+sin{45}^{\circ }\left)\right]$

or, $B_2=\frac{2{\mu }_0{i}_2}{4\pi \left(a/2\right)}\times \frac{2}{\sqrt{2}}=\frac{\sqrt{2}{\mu }_0{i}_2}{\pi a}\odot$

The direction of $B_1\text{and}{B}_2$ are perpendicularly inward and outward to plane, using right hand thumc rule.

$\because i_1>i_2$

Thus, $B_1>B_2$

Net magnetic filed at $\text{O}$,

$B_{net}=B_1-B_2$

$(\because B_1\text{and}{B}_2\text{are in opposite direction})$

$\therefore B_{net}=\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$