Question

A curent is fowing in a straight conductor wre of length L. The magnetic induction at a point perpendicular distance $\frac{L}{4}$ from its centre will be -

• A. $\frac{4\mu d}{\sqrt{5}a!}$
• B. $\frac{\mu d}{2\pi }$
• C. $\frac{{\mu }_0}{\sqrt{2L}}$
• D. Zero

Answer 1

The correct option is A $\frac{4\mu d}{\sqrt{5}a!}$

$\text{Solution:-}$

$B=\frac{{\mu }_{0}i}{4\pi R}\left(sin{\theta }_1+sin{\theta }_2\right)$

$\therefore sin{\theta }_2=\frac{L//2}{\sqrt{(L//2{)}^2+(L//2{)}^2}}$=$\frac{1}{2(\sqrt{1//4+1//16})}$=$\frac{4}{2\sqrt{5}}$

$⟹B=\frac{{\mu }_{0}i}{4\pi R}\left(\frac{2\times 4}{2\sqrt{5}}\right)B$

=$\frac{4{\mu }_{0}i}{\sqrt{5}\pi R}$

$\text{Hence the correct option is A}$