A wire in the form of a square of side \( a \) carries a current \( i \). Then the magnetic induction at the center of the square wire is (Magnetic permeability of free space = \( \mu_0 \)):
(A)
\( \frac{\mu_0 i}{2 \pi a} \)
(B)
\( \frac{\mu_0 i \sqrt{2}}{\pi a} \)
(C)
\( \frac{2 \sqrt{2} \mu_0 i}{\pi a} \)
(D)
\( \frac{\mu_0 i}{\sqrt{2} \pi a} \)
Show answer & solution
Verified Answer & Solution Locked
Sign in to your free student account to reveal verified answers, instant hints, and detailed explanations.
The ratio of the magnetic field at the center of a current carrying circular wire and the magnetic field at the center of a square coil made from the same length of wire will be:
(A)
\( \frac{\pi^2}{4\sqrt{2}} \)
(B)
\( \frac{\pi^2}{8\sqrt{2}} \)
(C)
\( \frac{\pi}{2\sqrt{2}} \)
(D)
\( \frac{\pi}{4\sqrt{2}} \)
Show answer & solution
Verified Answer & Solution Locked
Sign in to your free student account to reveal verified answers, instant hints, and detailed explanations.