Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center due to one side of the square loop is $B_1=\frac{{\mu }_{0}I}{4\pi (a/2)}(\sin 45^{\circ }+\sin 45^{\circ })=\frac{{\mu }_{0}I}{\sqrt{2}\pi a}$. For four sides, $B_{total}=4B_1=\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}$. So A is correct. The magnetic field is directly proportional to the current, so B is correct. By the right-hand thumb rule, the magnetic field at the center is perpendicular to the plane of the loop, so C is correct. For a circular loop of perimeter $4a$, its radius is $R=\frac{4a}{2\pi }=\frac{2a}{\pi }$. The magnetic field at its center is $B_{circ}=\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_{0}I}{2(2a/\pi )}=\frac{\pi {\mu }_{0}I}{4a}$. Comparing $B_{square}\approx 0.900\frac{{\mu }_{0}I}{a}$ and $B_{circ}\approx 0.785\frac{{\mu }_{0}I}{a}$, we find $B_{square}>B_{circ}$. Thus, D is incorrect.