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 $\oint \mathrm { B } \cdot d \mathbf { s }$ 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 $\mu$T.m
B)75 $\mu$T. m
C)19 $\mu$T .m
D)57 $\mu$T . m
E)38 $\mu$T.m

Question

Quizplus · Accepted Answer

The line integral of the magnetic field around a closed path is equal to the current enclosed by that path multiplied by the permeability of free space (μ0).

In this case, the square path is centered on the rod and lies in a plane perpendicular to the axis of the rod, so it encloses the entire cross section of the rod. Therefore, the current enclosed by the path is equal to the total current in the rod, which is given as 60 A.

The permeability of free space is a constant value, μ0 = 4π×10^-7 Tm/A.

So the line integral is:

Line integral = (current enclosed) × (μ0)
Line integral = 60 A × 4π×10^-7 Tm/A
Line integral = 2.4×10^-5 Tm

However, we need to convert the units to μF1F1F10T F1F1F1S1F1F1F10 m, using the fact that 1 Tm = 1 μF1F1F10T F1F1F1S1F1F1F10 m.

Line integral = 2.4×10^-5 Tm = 2.4×10^-5 μF1F1F10T F1F1F1S1F1F1F10 m

Rounding to two significant figures, the answer is 19 F1F1F1S1F1F1F10T F1F1F1S1F1F1F10 m, which corresponds to choice C.

A Conducting Rod with a Square Cross Section (3 ∮B⋅ds\oint \mathrm { B } \cdot d \mathbf { s }∮B⋅ds

A Conducting Rod with a Square Cross Section (3 $\oint \mathrm { B } \cdot d \mathbf { s }$