A circular loop of wire of radius 6.0 cm has 30 turns and lies in the xy plane.It carries a current of 5 A in such a direction that the magnetic moment of the loop is along the x axis.The magnetic field on the x axis at x = 6.0 cm is approximately
A)19 $\mu$T
B)0.56 mT
C)0.11 mT
D)47 $\mu$T
E)0.88 mT

Question

Quizplus · Accepted Answer

The magnetic field on the axis of a circular loop is given by the formula $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$, where $I$ is the current, $R$ is the radius of the loop, $x$ is the distance from the center of the loop along the axis, and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{Tm/A}$). For a coil with $N$ turns, this formula becomes $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$. Given $N = 30$, $I = 5 \, \text{A}$, $R = 0.06 \, \text{m}$, and $x = 0.06 \, \text{m}$, substituting these values into the formula gives a magnetic field of approximately $0.56 \, \text{mT}$.

A Circular Loop of Wire of Radius 6 μ\muμ T B)056 MT C)0

A Circular Loop of Wire of Radius 6 $\mu$ T
B)056 MT
C)0