A conducting rod with a square cross section (3.0 cm × 3.0 cm)carries a current of 60 A that is uniformly distributed across the cross section.What is the magnitude of the (line)integral
Around a square path (1.5 cm × 1.5 cm)if the path is centered on the center of the rod and lies in a plane perpendicular to the axis of the rod?
A)14 μT ⋅ m
B)75 μT ⋅ m
C)19 μT ⋅ m
D)57 μT ⋅ m
E)38 μT ⋅ m

Question

A conducting rod with a square cross section (3.0 cm × 3.0 cm)carries a current of 60 A that is uniformly distributed across the cross section.What is the magnitude of the (line)integral 
Around a square path (1.5 cm × 1.5 cm)if the path is centered on the center of the rod and lies in a plane perpendicular to the axis of the rod?
A)14 μT ⋅ m 
B)75 μT ⋅ m 
C)19 μT ⋅ m 
D)57 μT ⋅ m 
E)38 μT ⋅ m

Quizplus · Accepted Answer

The magnetic field at a point on the square path due to a small element of length dl carrying current I is given by:
$\delta B=\frac{\mu_0}{4\pi} \frac{Idl	imes\hat{r}}{r^2}$
where $\hat{r}$ is the unit vector in the radial direction, r is the distance from the element to the point on the square path, and $\mu_0$ is the permeability of free space. The total magnetic field at the point on the square path due to the current in the entire conducting rod is found by integrating over all the small elements of the rod:
$B=\int \delta B$
The magnetic field is perpendicular to both the current direction and the radial direction, so it points in the tangential direction around the path. The magnetic field has the same magnitude at every point on the path because of the symmetry of the problem.
Using the fact that the magnetic field is perpendicular to the square path, we can simplify the integral by considering only the top and bottom sides of the square path. The integral around each of these sides is the same because of the symmetry of the problem. We have:
$B=4\int_{0}^{1.5	ext{ cm}} \delta B=\frac{4\mu_0 I}{4\pi}\int_{0}^{1.5	ext{ cm}} \frac{dx}{(d^2/4 + x^2)^{3/2}}$
where d is the diagonal of the conducting rod, which is $\sqrt{3}	imes 3	ext{ cm}=5.2	ext{ cm}$. Evaluating the integral gives:
$B=\frac{\mu_0 I d^2}{2\pi}\left(\frac{1}{d^2/4}-\frac{1}{d^2/4 + (1.5	ext{ cm})^2}ight)\approx 19\mu	ext{T}\cdot	ext{m}$

A Conducting Rod with a Square Cross Section (3