Search any question & find its solution
Question:
Answered & Verified by Expert
A current I flows in the anticlockwise direction through a square loop of side a lying in the xoy plane with its center at the origin. The magnetic induction at the center of the square loop is
Options:
Solution:
1438 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \sqrt{2} \mu_{0} I}{\pi a} \hat{e}_{z}$
Field due to one side of loop at $O$
$$
=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left(2 \sin 45^{\circ}\right)
$$
Field at $O$ due to all four sides is along unit vector $\hat{e}_{z}$
$\therefore$ Total field
$=4 \cdot \frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left(2 \sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
$$
=\frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left(2 \sin 45^{\circ}\right)
$$
Field at $O$ due to all four sides is along unit vector $\hat{e}_{z}$
$\therefore$ Total field
$=4 \cdot \frac{\mu_{0} I}{4 \pi\left(\frac{a}{2}\right)}\left(2 \sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.