Question

An electric current of I = 6A flows in a conductor bent into a square ABCD with side a = 10cm. Determine the magnetic induction vector B and the magnetic field strength at the center O of the circuit.

Answer 1

$|B|=4B_1,$

$B_1=\frac{\mu _0 I}{4\pi \frac a2}(cos 45°-cos 135°)=\frac{\mu _0 I \sqrt2}{2\pi a}=17 ~\mu T,$

$B=68~\mu T.$

$H=\frac{B}{\mu _0}=\frac{68\cdot 10^{-6}}{4 \pi \cdot 10^{-7}}=54 ~Oe.$