Question

As shown in the figure, a long, hollow, conducting cylinder of inner radius \(a\) and outer radius \(b\) carries a current that is flowing out of the page. Suppose that \(a=5.00 \mathrm{~cm}, b=7.00 \mathrm{~cm},\) and the current \(i=100 . \mathrm{mA}\) uniformly distributed over the cylinder wall (between \(a\) and \(b\) ). Find the magnitude of the magnetic field at each of the following distances \(r\) from the center of the cylinder: a) \(r=4.00 \mathrm{~cm}\) b) \(r=6.50 \mathrm{~cm}\) c) \(r=9.00 \mathrm{~cm}\)

Short answer

Answer: The magnetic field magnitudes at distances r=4.00 cm, r=6.50 cm, and r=9.00 cm are 0 T, 1.36e-4 T, and 4.44e-5 T, respectively.

Step-by-step solution

Ampere's Law

Ampere's law for magnetic fields states that the magnetic field integrated along a closed path is equal to μ₀ times the current enclosed by that path. Mathematically, this is $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$$

Define Coordinates and l

Let's define a cylindrical coordinate system, where the z-axis is aligned with the axis of the cylinder. We want to find the magnetic field magnitude, |B|, along a circle of radius r centered at the origin. The magnetic field will be tangent to this circle, and the path length along this circle is $$d\vec{l} = r\,d\varphi\,\hat\varphi$$

Calculate Enclosed Current

To find the magnetic field at a particular distance r, we need to find the current enclosed by the circle of that radius. The total current, i=100mA, is uniformly distributed over the cylinder wall between inner radius a and outer radius b. The current has the density $$J = \frac{i}{2\pi(b-a)h}$$ where h is the height of the cylinder. Define the enclosed current as I(r). Then, for the different distances r we have: - r < a: I(r) = 0 - a≤r≤b: I(r) = J*2πr*h - r > b: I(r) = i

Ampere's Law for Different Distances

Using Ampere's Law and the enclosed currents derived above, we obtain expressions for the magnetic field magnitude |B| at different distances r: 1. r < a: $$\oint \vec{B}\cdot d\vec{l} = 0$$ so |B| = 0. 2. a≤r≤b: $$\oint \vec{B}\cdot d\vec{l} = \mu_0 J*2\pi r h$$ so $$|B| = \frac{\mu_0 J r}{(b - a)}$$ 3. r > b: $$\oint \vec{B}\cdot d\vec{l} = \mu_0 i$$ so $$|B| = \frac{\mu_0 i}{2\pi r}$$

Calculate Magnetic Field for Given Distances

Now we can compute the magnetic field magnitude for the given distances, using the appropriate formula for each case: a) r = 4.00 cm Since r < a, the magnetic field magnitude is 0 T. b) r = 6.50 cm Since a≤r≤b, we use the expression for |B| between inner and outer radius: $$|B| = \frac{\mu_0 J r}{(b - a)}$$ with r=6.50 cm, a=5.00 cm, b=7.00 cm, i=100 mA and $$\mu_0 = 4\pi \times10^{-7} Tm/A$$ we find |B| = 1.36e-4 T. c) r = 9.00 cm Since r > b, we use the expression for |B| outside the cylinder: $$|B| = \frac{\mu_0 i}{2\pi r}$$ with r=9.00 cm, i=100 mA and $$\mu_0 = 4\pi \times10^{-7} Tm/A$$ we find |B| = 4.44e-5 T. So, the magnitudes of the magnetic field at distances r=4.00 cm, r=6.50 cm, and r=9.00 cm are 0 T, 1.36e-4 T, and 4.44e-5 T, respectively.