Question

Find magnetic filed at the centre of the rectangle as shown in the figure.

• A. $\frac{4{\mu }_{0}I}{\pi ab}\sqrt{b^2+a^2}$
• B. $\frac{2{\mu }_{0}I}{\pi a{b}^2}\sqrt{b^2+a^2}$
• C. $\frac{2{\mu }_{0}I}{\pi ab}\sqrt{b^2+a^2}$
• D. $\frac{2{\mu }_{0}I}{\pi b{a}^2}\sqrt{b^2+a^2}$

Answer 1

The correct option is C $\frac{2{\mu }_{0}I}{\pi ab}\sqrt{b^2+a^2}$

For the given rectangular loop $\text{ABCD}$, fields due to side $\text{AD}$ and $\text{BC}$ will be equal. Similarly fields due side $\text{AB}$ and $\text{CD}$ will be equal.


Magnetic field due to side $\text{A}D$ is ,

$\because$ Point $\text{O}$ is on the perpendicular bisector of side $\text{A}D$, field at $\text{O}$ will be,

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

$B_{AD}=\frac{{\mu }_{0}I}{4\pi d}\left(2\sin \theta \right).........\left(1\right)[\because \sin {\theta }_1=\sin {\theta }_2=\sin \theta )]$

From the $△AOH$ ,

$\sin \theta =\frac{\frac{b}{2}}{\sqrt{{\left(\frac{b}{2}\right)}^2+(d{)}^2}}[\because d=\frac{a}{2}]$

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

From (1) we get,

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

Similarly, magnetic field due to side $\text{C}D$ is ,


From the $△DOH$ ,

$\sin \theta =\frac{\frac{a}{2}}{\sqrt{{\left(\frac{a}{2}\right)}^2+(d{)}^2}}[\because d=\frac{b}{2}]$

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

From (1) we get,

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

Net field at the centre of the rectangle is,

$B_{\text{net}}=2B_{AD}+2B_{CD}⨂$

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

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

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

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.