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?

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?

Quizplus · Accepted Answer

According to Ampere’s Law, the magnitude of the line integral of the magnetic field around a closed loop is proportional to the current passing through the loop. We can use this law to find the magnetic field along the square path.

The current passing through the square path is the same as the current passing through the rod, since the current is uniformly distributed across the cross section. Therefore, we can use the formula

B = μ0I / (2a)

where B is the magnetic field, μ0 is the permeability of free space, I is the current, and a is the distance from the center of the rod to the center of the square path.

The distance a is half the distance between opposite faces of the rod, so a = 3.0 cm / 2 = 1.5 cm.

Plugging in the given values, we get

B = (4π×10^-7 T·m/A)(60 A) / (2×0.015 m) = 0.19 T = 1900 μT

The line integral is the product of the magnetic field and the length of the path, which is 4 times the length of one side of the square path. Therefore,

Line integral = 4BL = 4(1900 μT)(0.015 m) = 114 μT·m ≈ 19 μT·m

So the answer is (C).

A Conducting Rod with a Square Cross Section (3