Question

$A$ $4A$ current carrying loop consists of three identical quarter circles of radius $5cm$ lying in the positive quadrants of the $x-y$ $y-z$ and $z-x$ planes with their centres at the origin joined together, value of $\vec{B}$ at the origin is

• A. $\frac{{\mu }_0}{10}\left(\hat{i}+\hat{j}-\hat{k}\right)T$
• B. $\frac{{\mu }_0}{10}\left(-\hat{i}+\hat{j}+\hat{k}\right)T$
• C. $\frac{{\mu }_0}{5}\left(\hat{i}+\hat{j}+\hat{k}\right)T$
• D. $10{\mu }_0\left(\hat{i}+\hat{j}+\hat{k}\right)T$

Answer 1

Answer: (D)

As $\vec{B}={\vec{B}}_{xy}+{\vec{B}}_{yz}+{\vec{B}}_{zx}\dots \left(i\right)$
where ${\vec{B}}_{xy}=\frac{{\mu }_0}{4\pi }\frac{I}{R}\theta \hat{k}$,
${\vec{B}}_{yz}=\frac{{\mu }_0}{4\pi }\frac{I}{R}\theta \hat{i}$,
${\vec{B}}_{zx}=\frac{{\mu }_0}{4\pi }\frac{I}{R}\theta \hat{j}$,
Substituting these values in equation $\left(i\right)$ we get,
$\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I}{R}\theta \left[\hat{i}+\hat{j}+\hat{k}\right]$
Here, $\theta =\frac{\pi }{2},I=4A$ and $R=5cm=5\times 10^{-2}m$
$\therefore \vec{B}=\frac{{\mu }_0}{4\pi }\times \frac{4}{5\times 10^{-2}}\times \frac{\pi }{2}\left(\hat{i}+\hat{j}+\hat{k}\right)$
$=10{\mu }_0\left(\hat{i}+\hat{j}+\hat{k}\right)T$