Question

Find the net magnetic filed at the centre $\left(O\right)$ of the rectangle.

• A. $\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{ab}\right)$
• B. $\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{a{b}^2}\right)$
• C. $\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{a^{2}b}\right)$
• D. $\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{(ab{)}^2}\right)$

Answer 1

The correct option is A $\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{ab}\right)$

For the given rectangle, side $\text{AB}$ and $\text{CD}$ are equal in length and carries same current, hence they will produce same field at the centre of the loop. Similarly, side $\text{BC}$ and $\text{DA}$ will produce same field at the centre of the loop.

Let $B_1$ be the field due to side $\text{DA}$.


The magnetic field due to a finite wire is,

$B_1=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

Here, $\sin {\theta }_1=\sin {\theta }_2=\sin \theta$

$\therefore B_1=\frac{{\mu }_{0}I}{2\pi d}\left(\sin \theta \right)...\left(1\right)$

Also, $\sin \theta =\frac{b/2}{\left(\sqrt{a^2+b^2}\right)/2}$

$\Rightarrow \sin \theta =\frac{b}{\sqrt{a^2+b^2}}$

From $\left(1\right)$, we get,

$B_1=\frac{{\mu }_{0}I}{2\pi d}\frac{b}{\sqrt{a^2+b^2}}=\frac{{\mu }_{0}I}{2\pi \left(a/2\right)}\frac{b}{\sqrt{a^2+b^2}}$

$\therefore B_1=\frac{{\mu }_{0}I}{\pi a}\frac{b}{\sqrt{a^2+b^2}}$

Similarly, magnetic field due to side $\text{AB}$,

$B_2=\frac{{\mu }_{0}I}{\pi b}\frac{a}{\sqrt{a^2+b^2}}$

The resultant field due to all four sides is,

$B_{net}=2B_1+2B_2=2(B_1+B_2)$

$=2[\frac{{\mu }_{0}I}{\pi a}\frac{b}{\sqrt{a^2+b^2}}+\frac{{\mu }_{0}I}{\pi b}\frac{a}{\sqrt{a^2+b^2}}]$

$=\frac{2{\mu }_{0}I}{\pi \sqrt{a^2+b^2}}(\frac{b}{a}+\frac{a}{b})$

$=\frac{2{\mu }_{0}I}{\pi \sqrt{b^2+a^2}}\left(\frac{b^2+a^2}{ab}\right)$

$=\frac{2{\mu }_{0}I}{\pi }\left(\frac{\sqrt{b^2+a^2}}{ab}\right)$