A conducting hollow cylinder (inner radius = a,outer radius = b)carries a current of 40 A that is uniformly distributed over the cross section of the conductor.If a = 3.0 mm and b = 6.0 mm,what is the magnitude of the (line)integral $\oint \mathrm { B } \cdot d \mathbf { s }$ around a circular path (radius = 5.0 mm)centered on the axis of the cylinder and in a plane perpendicular to that axis?
A)50 $\mu$T . m
B)30 $\mu$T . m
C)22 $\mu$T. m
D)37 $\mu$T . m
E)47$\mu$T . m

Question

Quizplus · Accepted Answer

The given line integral represents the magnetic field due to the current-carrying cylinder at a distance of 5.0 mm from the axis and in a plane perpendicular to it. Using Ampere's law and considering a circular path of radius 5.0 mm, we can write:

∫B.dl = μ0 I_enc

where μ0 is the permeability of free space and I_enc is the current enclosed by the circular path.

For a point inside the cylinder (r < a), the enclosed current is 0, and hence the magnetic field is 0. For a point outside the cylinder (r > b), the enclosed current is equal to the total current (40 A), and hence we can use Ampere's law directly to find the magnetic field:

B = μ0 I_enc / (2πr) = (μ0/2π) * (40 / (π(b^2 - a^2))) = 4μ0 / ((b^2 - a^2))

Putting in the given values, we get:

B = 4π × 10^-7 Tm/A * 40 A / (2π*5.0 × 10^-3 m) = 8 × 10^-4 T

Finally, the line integral of B around the circular path is simply the product of B and the circumference of the circle:

∫B.dl = B * 2πr = 8 × 10^-4 T * 2π * 5.0 × 10^-3 m = 0.03 μT m

Therefore, the answer is B.

A Conducting Hollow Cylinder (Inner Radius = A,outer Radius = ∮B⋅ds\oint \mathrm { B } \cdot d \mathbf { s }∮B⋅ds

A Conducting Hollow Cylinder (Inner Radius = A,outer Radius = $\oint \mathrm { B } \cdot d \mathbf { s }$