Question

$A$ $4\, A $ current carrying loop consists of three identical quarter circles of radius $5 \,cm$ 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 $\vec{B}=\vec{B}_{xy}+\vec{B}_{yz}+\vec{B}_{zx} \ldots\left(i\right)$
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, $\theta=\frac{\pi}{2}, I=4\,A$ and $R=5\,cm=5\times10^{-2}\,m$
$\therefore \vec{B}=\frac{\mu_{0}}{4\pi}\times\frac{4}{5\times10^{-2}}\times\frac{\pi}{2} \left(\hat{i}+\hat{j}+\hat{k}\right)$
$=10\mu_{0}\left(\hat{i}+\hat{j}+\hat{k}\right)T$