Question

The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of a/2 from its centre (as shown is)

• A. $\frac{\sqrt{{2\mu }_{0}i}}{\sqrt{3\pi a}}$
• B. $\frac{{\mu }_{0}i}{\sqrt{6\pi a}}$
• C. $\frac{{2\mu }_{0}i}{\sqrt{3}\pi a}$
• D. $zero$

Answer 1

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

$Given:$ Side=$a$ ; distance=$\frac{a}{2}$

$Solution:$

$B_{\text{net}}=4B\cos {45}^{\circ }=\frac{4}{\sqrt{2}}B$

$B_{\text{net}}=2\sqrt{2}B$ $.........\left(1\right)$

From Fig

$\sin \alpha =\frac{a/2}{\frac{\sqrt{3}a}{2}}=\frac{1}{\sqrt{3}}$

$$\begin{array}{cc}d& =\frac{a}{\sqrt{2}}\\ \text{But},B& =\frac{{\mu }_{0}i}{4\pi d}(\sin \alpha +\sin \beta )\\ & =\frac{{\mu }_{o}i}{4\pi a_{/\sqrt{2}}}\left(2\sin \alpha \right)\end{array}$$

$B=\frac{\mu 0i\sqrt{2}}{4\pi a}\left({}^2/\sqrt{3}\right).........\left(2\right)$

$Equation\left(2\right)in\left(1\right)$

$B_{net}$ $=\frac{\text{/}4}{\text{/}\sqrt{2}}\times \frac{{\mu }_{0}i2\text{/}\sqrt{2}}{\text{/}4\pi \sqrt{3}}$

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

$So,the$ $correct$ $option:C$