Search any question & find its solution
Question:
Answered & Verified by Expert
A square wire of each side $l$ carries a current $I$. The magnetic field at the mid-point of the square
Options:
Solution:
1861 Upvotes
Verified Answer
The correct answer is:
$8 \sqrt{2} \frac{\mu_0}{4 \pi} \frac{I}{l}$
$B=4\left[\frac{\mu_0}{4 \pi} \cdot \frac{I}{a}\left(\sin \phi_1+\sin \phi_2\right)\right]$
Here, $a=\frac{l}{2}$ and $\phi_1=\phi_2=45^{\circ}$
$\begin{aligned}
B & =4 \times \frac{\mu_0}{4 \pi} \frac{I}{(l / 2)}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \\
\therefore \quad & B=\frac{16}{\sqrt{2}}\left[\frac{\mu_0}{4 \pi} \frac{I}{l}\right]=8 \sqrt{2}\left(\frac{\mu_0}{4 \pi} \cdot \frac{I}{l}\right)
\end{aligned}$
Here, $a=\frac{l}{2}$ and $\phi_1=\phi_2=45^{\circ}$
$\begin{aligned}
B & =4 \times \frac{\mu_0}{4 \pi} \frac{I}{(l / 2)}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \\
\therefore \quad & B=\frac{16}{\sqrt{2}}\left[\frac{\mu_0}{4 \pi} \frac{I}{l}\right]=8 \sqrt{2}\left(\frac{\mu_0}{4 \pi} \cdot \frac{I}{l}\right)
\end{aligned}$
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.