A long hollow cylindrical conductor (inner radius = 2.0 mm, outer radius = 4.0 mm) carries a current of 24 A distributed uniformly across its cross section. A long wire which is coaxial with the cylinder carries an equal current in the opposite direction. What is the magnitude of the magnetic field 3.0 mm from the axis?

Question

Quizplus · Accepted Answer

First, we need to find the magnetic field due to the hollow cylindrical conductor alone. Using Ampere's law, we can find the magnetic field inside the cylinder:
$$B_1 = \frac{\mu_0 I}{2\pi r}$$
where $\mu_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance from the axis. Since the current is distributed uniformly across the cross section, we can use the current per unit length $J$ as follows:
$$I = J \cdot A$$
where $A$ is the cross-sectional area of the cylinder, given by:
$$A = \pi (r_2^2 - r_1^2)$$
where $r_1$ and $r_2$ are the inner and outer radii, respectively. Thus:
$$I = J \cdot \pi (r_2^2 - r_1^2)$$
Plugging this into the equation for $B_1$ and simplifying:
$$B_1 = \frac{\mu_0 J}{2} \cdot \frac{r_2^2 - r_1^2}{r}$$
Now, we can use the principle of superposition to find the magnetic field at a distance of 3.0 mm from the axis due to both the cylinder and the wire. Since the wire carries an equal current in the opposite direction, its contribution to the magnetic field is equal and opposite to that of the cylinder. Therefore, the total magnetic field at a distance of 3.0 mm from the axis is simply the difference between the magnetic field due to the cylinder alone and that due to the cylinder and wire together:
$$B = B_1 - B_2$$
where $B_2$ is the magnetic field due to the wire alone, given by:
$$B_2 = \frac{\mu_0 I}{2\pi r}$$
Plugging in the values for $r_1$, $r_2$, $I$, $r$, and $\mu_0$, we get:
$$B_1 = \frac{2 	imes 10^{-7} \cdot (24/(\pi \cdot (0.004^2 - 0.002^2)))}{2\pi \cdot 0.003} \approx 1.87 	ext{ mT}$$
$$B_2 = \frac{2 	imes 10^{-7} \cdot 24}{2\pi \cdot 0.003} \approx 1.46 	ext{ mT}$$
Therefore, the total magnetic field is:
$$B \approx 0.41 	ext{ mT}$$
which is closest to option (B) 0.93 mT.

A Long Hollow Cylindrical Conductor (Inner Radius = 2