Search any question & find its solution
Question:
Answered & Verified by Expert
As shown in the figure, a single conducting wire is bent to form a loop in the form of a circle of radius 'r' concentrically inside a square of side ' $a$ ', where $a: r=8: \pi$. A battery B drives a current through the wire. If the battery B and the gap G are of negligible sizes, determine the strength of magnetic field at the common centre $O$.
Solution:
2531 Upvotes
Verified Answer
The correct answer is:
$\frac{\mu_{0} I}{\pi \mathrm{a}} 2 \sqrt{2}(\sqrt{2}-1)$
Hint:
$\frac{a}{r}=\frac{8}{\pi}, r=\frac{\pi a}{8}$
$B=\frac{\mu_{0} \mid}{2 r}-\frac{4 \mu_{0} I}{4 \pi a / 2} \times \sqrt{2}=\frac{\mu_{0} I}{2 r}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}=\frac{\mu_{0} \mid \times 8}{2 \pi a}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
$=\frac{\mu_{0} \mid}{\pi a}(4-2 \sqrt{2})=\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}-1)$
$\frac{a}{r}=\frac{8}{\pi}, r=\frac{\pi a}{8}$
$B=\frac{\mu_{0} \mid}{2 r}-\frac{4 \mu_{0} I}{4 \pi a / 2} \times \sqrt{2}=\frac{\mu_{0} I}{2 r}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}=\frac{\mu_{0} \mid \times 8}{2 \pi a}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
$=\frac{\mu_{0} \mid}{\pi a}(4-2 \sqrt{2})=\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}-1)$
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.