Question

A square loop of edge $a=20\text{cm}$ made of uniform wire is as shown in the figure below. If current $i=5\text{A}$ is entering into the system at $\text{A}$ and leaves at $\text{C}$. Find the magnetic field (approximately) at point $\text{P}$ which is on perpendicular bisector of $\text{AB}$ (inside the loop) at a distance $5\text{cm}$ from it.

• A. $7\mu \text{T}⨀$
• B. $70\mu \text{T}⨀$
• C. $0.7\mu \text{T}⨀$
• D. $0.07\mu \text{T}⨀$

Answer 1

The correct option is A $7\mu \text{T}⨀$

Current distribution in each wire is as shown in the figure below.


Due to symmetry, magnetic field due to both $5\text{A}$ current element will sum to zero.

Net magnetic field due to the system of wires at point $P$ is $\overrightarrow{B}=\overrightarrow{B_{\text{AB}}}+\overrightarrow{B_{\text{BC}}}+\overrightarrow{B_{\text{DC}}}+\overrightarrow{B_{\text{AD}}}$

Magnetic field due to $AD$ and $BC$ will be equal but opposite (due to symmetry). So $\overrightarrow{B_{\text{BC}}}+\overrightarrow{B_{\text{AD}}}=0$

Now, $B_{\text{AB}}=\frac{{\mu }_0\times 2.5}{4\pi \times 5\times {10}^{-2}}[\frac{\left(10\right)}{\sqrt{{\left(5\right)}^2+{\left(10\right)}^2}}+\frac{\left(10\right)}{\sqrt{{\left(5\right)}^2+{\left(10\right)}^2}}]$

$B_{\text{AB}}=\frac{{\mu }_0}{2\pi \sqrt{5}\times {10}^{-2}}⨀$

$B_{\text{CD}}=\frac{{\mu }_0\times 2.5}{4\pi \times 15\times {10}^{-2}}[\frac{\left(10\right)}{\sqrt{{\left(15\right)}^2+{\left(10\right)}^2}}+\frac{\left(10\right)}{\sqrt{{\left(15\right)}^2+{\left(10\right)}^2}}]$

$B_{\text{CD}}=\frac{{\mu }_0}{6\pi \sqrt{13}\times {10}^{-2}}⨂$

$\overrightarrow{B}=\frac{{\mu }_0}{2\pi \times {10}^{-2}}(\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}})$

Substituting the given values we get,

$\overrightarrow{B}=\frac{4\pi \times {10}^{-7}}{2\pi \times {10}^{-2}}(\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}})⨀$

$\overrightarrow{B}=2\times {10}^{-5}\times (0.44-0.092)=7\times {10}^{-6}\text{T}⨀$

Hence, option (a) is the correct answer.