A square loop of wire,20 cm long on each side carries a current of 0.3 A.What is the magnetic field in the center of the square loop?
A)0.5 µT
B)1.2 µT
C)1.9 µT
D)2.3 µT

Question

Quizplus · Accepted Answer

The magnetic field at the center of a square loop can be calculated using the formula for the magnetic field at the center of a straight wire segment bent into an arc of a circle. For a square loop, this is effectively calculated as the field due to four such arcs, with the formula $B = \frac{{\mu_0 I}}{{2\pi r}}(\phi)$, where $\phi$ is the angle subtended by the arc (in radians), $I$ is the current, $r$ is the radius (or half the length of the side of the square in this case, since the field is being calculated at the center), and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{Tm/A}$). For a square loop, each side can be considered as part of a circle that subtends a right angle at the center, so $\phi = \frac{\pi}{2}$ for each side, and there are four sides, making the total angle $4 \times \frac{\pi}{2} = 2\pi$. Thus, the formula simplifies to $B = \frac{{\mu_0 I}}{{2r}}$, with $r = 0.1$ m (half the length of the side of the square). Plugging in the values, $B = \frac{{4\pi \times 10^{-7} \times 0.3}}{{2 \times 0.1}} = 1.9 \times 10^{-6}$ T, or 1.9 µT.

A Square Loop of Wire,20 Cm Long on Each Side