A long, coaxial cable, shown in cross-section in the drawing, is made using two conductors that share a common central axis, labeled C. The conductors are separated by an electrically insulating material that is also used as the outer cover of the cable. The current in the inner conductor is 2.0 A directed into the page and that in the outer conductor is 2.5 A directed out of the page. The distance from point C to point A is 0.0015 m; and the distance from C to B is 0.0030 m. The radii a and b of the conductors are 6.0 × 10-4 m and 1.9 × 10-3 m, respectively.
-What is the magnitude and direction of the magnetic field at point B?

Question

A long, coaxial cable, shown in cross-section in the drawing, is made using two conductors that share a common central axis, labeled C. The conductors are separated by an electrically insulating material that is also used as the outer cover of the cable. The current in the inner conductor is 2.0 A directed into the page and that in the outer conductor is 2.5 A directed out of the page. The distance from point C to point A is 0.0015 m; and the distance from C to B is 0.0030 m. The radii a and b of the conductors are 6.0 × 10-4 m and 1.9 × 10-3 m, respectively. 
-What is the magnitude and direction of the magnetic field at point B?

Quizplus · Accepted Answer

The magnetic field due to a current-carrying wire is given by the Biot-Savart law:$$dB = \frac{\mu_0}{4\pi} \frac{Idl imes \hat{r}}{r^2}$$where:* dB is the magnetic field vector* μ0 is the vacuum permeability* I is the current* dl is the vector representing the length of the wire segment* r is the vector representing the distance from the wire segment to the point where the magnetic field is being calculated* ^r is the unit vector in the direction of rIn this case, we have a wire carrying a current of 5 A, and we want to find the magnetic field at a point that is 0.1 m from the wire. We can use the Biot-Savart law to calculate the magnetic field:$$dB = \frac{\mu_0}{4\pi} \frac{Idl imes \hat{r}}{r^2}$$$$dB = \frac{(4\pi imes 10^{-7} ext{ T$\cdot$m/A})(5 ext{ A})(0.1 ext{ m}) \hat{z}}{0.1^2 ext{ m}^2}$$$$dB = 3.3 imes 10^{-5} ext{ T}$$The direction of the magnetic field is given by the cross product of dl and ^r. In this case, dl is in the positive y-direction and ^r is in the positive x-direction, so the cross product is in the positive z-direction. Therefore, the magnetic field is in the counterclockwise direction.The correct answer is B) 3.3 × 10^-5 T, counterclockwise.

A Long, Coaxial Cable, Shown in Cross-Section in the Drawing