Question

A long, straight, nonmagnetic conductor of \(0.2 \mathrm{~mm}\) radius carries a uniformly distributed current of 2 A dc. \((a)\) Find \(J\) within the conductor. (b) Use Ampère's circuital law to find \(\mathbf{H}\) and \(\mathbf{B}\) within the conductor. (c) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) within the conductor. \((d)\) Find \(\mathbf{H}\) and \(\mathbf{B}\) outside the conductor. \((e)\) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) outside the conductor.

Short answer

Question: Calculate the current density within a non-magnetic conductor, the magnetic field and magnetic induction both inside and outside the conductor, and demonstrate that the curl of the magnetic field is equal to the current density in both cases. Answer: The current density within the conductor is \(15.9\times 10^{6} \,\text{A/m}^2\). The magnetic field and magnetic induction inside the conductor are given by \(H(r) = 7.95\times10^6 r \,\text{A/m}\) and \(B(r) = 10^{-7} \pi 7.95\times10^6 r \,\text{T}\), respectively. Outside the conductor, they are given by \(H(r) = \dfrac{2\,\text{A}}{2\pi r}\) and \(B(r) = 10^{-7} 2\pi \dfrac{2\,\text{A}}{r}\,\text{T}\), respectively. The curl of the magnetic field is equal to the current density both inside and outside the conductor, as shown in the calculations.

Step-by-step solution

(a) Current density calculation

The current density (\(J\)) is equal to the total current divided by the cross-sectional area of the conductor: \(J = \dfrac{I}{A}= \dfrac{2\,\text{A}}{(\pi r^2)}\), where \(r=0.2\,\text{mm}=0.0002\,\text{m}\). Now we can find \(J\): \(J = \dfrac{2\,\text{A}}{(\pi (0.0002)^2)} = 15.9\times 10^{6} \,\text{A/m}^2\).

(b) Calculate \(\mathbf{H}\) and \(\mathbf{B}\) inside the conductor

We will use Ampère's circuital law to find the magnetic field inside the conductor. Ampère's circuital law can be expressed as: \(\oint \mathbf{H} \cdot d\mathbf{L} = I_{\text{enc}}\), where \(I_{\text{enc}}\) is the enclosed current by a closed loop. Choose a circular path with radius \(r < 0.2\,\text{mm}\) concentric with the conductor. Due to symmetry, the magnetic field will be equal at every point along this loop and only its azimuthal component will be non-zero. We can rewrite the Ampère's law for this loop: \(H(r) \cdot 2\pi r = I_{\text{enc}}\), \(I_{\text{enc}} = J \cdot A_{\text{enc}} = J \pi r^2\). Solving for \(H(r)\) we obtain: \(H(r) = \dfrac{J \cdot r}{2} = 7.95\times10^6 r \,\text{A/m}\). The magnetic induction inside the conductor can be calculated from \(B = \mu_0 H\): \(B(r) = \mu_0 H(r) = 10^{-7} \pi 7.95\times10^6 r \,\text{T}\).

(c) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) inside the conductor

We can compute the curl of the magnetic field: \(\nabla \times \mathbf{H} = (\frac{\partial}{\partial r}) (7.95\times 10^6)\hat{\theta}\). Computing the curl gives us \(\nabla \times \mathbf{H} = 7.95\times 10^6 \,\text{A/m}^2\hat{z}\), which is equal to the current density \(\mathbf{J} = J \hat{z}\) that we computed in (a).

(d) Find \(\mathbf{H}\) and \(\mathbf{B}\) outside the conductor

We apply Ampère's circuital law to a circular loop of radius \(r > 0.2\,\text{mm}\) concentric with the conductor. The enclosed current \(I_{\text{enc}}\) is equal to the total current, i.e. \(I_{\text{enc}} = 2 \text{A}\). Using the same approach as for the inside case, we obtain: \(H(r) = \dfrac{I}{2 \pi r} = \dfrac{2\,\text{A}}{2\pi r}\). The magnetic induction outside the conductor can be calculated from \(B = \mu_0 H\): \(B(r) = \mu_0 H(r) = 10^{-7} 2\pi \dfrac{2\,\text{A}}{r}\,\text{T}\).

(e) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) outside the conductor

The curl of the magnetic field outside the conductor is: \(\nabla \times \mathbf{H} = (\frac{\partial}{\partial r}) (10^{-6} \dfrac{1}{r})\hat{\theta}\). When computing the curl, we obtain \(\nabla \times \mathbf{H} = 0\), which is equal to the current density \(\mathbf{J} = 0 \hat{z}\) outside the conductor since there is no current.

Key concepts

Current Density

Imagine a group of water molecules flowing through a pipe. The amount of water that passes along every second is pretty much the same as the concept of 'current density' in the world of electricity. It's a measure of how much electric charge is flowing through a conductor's cross-sectional area per second. To find the current density, designated by the symbol \(J\), we simply divide the total current \(I\), measured in amperes (A), by the area \(A\) in square meters (\(\text{m}^2\)).

Mathematically, the current density formula appears as: \[ J = \frac{I}{A} \]. In our exercise, we're dealing with a long, straight conductor carrying a 2A direct current (dc). The conductor itself has a radius of 0.2 mm, which means its cross-sectional area is a small circle. By dividing the total current by this tiny circular area, we determine how 'densely' the current is packed within the conductor.

Magnetic Field

Now, picture that our water in the pipe is surrounded by an invisible force. In electricity, when current flows through a conductor, it's surrounded by a 'magnetic field', a fundamental force in nature that exerts a push or pull on moving charges. It's invisible to the naked eye, but it's there, and it can be felt by other magnetic objects.

The intensity of this magnetic field is typically represented by the symbol \(\textbf{H}\), and is measured in amperes per meter (\(A/m\)). Using Ampère's circuital law, which bridges the relationship between current and magnetic field, we can compute \(\textbf{H}\) by integrating the magnetic field around a closed loop. This law can be given as: \[ \oint \textbf{H} \cdot d\textbf{L} = I_{\text{enc}} \], where \(I_{\text{enc}}\) is the current enclosed by the loop. Inside the conductor, \(H\) varies with distance from its center, leading to a different magnetic field at different points within the conductor.

Magnetic Induction

If you've ever played with magnets, you've seen how they can cause other magnetic objects to move without touching them. This phenomenon can be explained by 'magnetic induction'. Magnetic induction refers to the generation of a magnetic field around an electrical conductor as electric current flows through it.

In our exercise, magnetic induction is represented as \(\textbf{B}\), known as the magnetic flux density and is measured in tesla (T). The relationship between the magnetic field \(H\) and magnetic induction \(B\) is linked by \(B = \mu_0 H\), where \(\mu_0\) is the vacuum permeability, a constant value that shows how well a vacuum can support the formation of a magnetic field. So when students are calculating \(B\), they're essentially taking the magnetic field strength \(H\) and figuring out how that translates into the actual magnetic 'force' at a point in space.

Maxwell's Equations

Imagine a universe where electricity, magnetism, light, and even radio waves are all connected by a set of four incredible rules. These rules are 'Maxwell's equations', named after the scientist James Clerk Maxwell, who first penned them. They are like the playbook for how electric and magnetic fields behave and how they relate to charges and currents.

Ampère’s circuital law, used in our exercise, is actually one of these equations - and specifically, it's the one that tells us about the relationship between magnetic fields and the currents that produce them. In the context of our problem, Maxwell's equations hint at a deeper principle behind part (c) and (e): \( abla \times \textbf{H}=\textbf{J} \) within the conductor and just why the curl (\( abla \times \)) of \(\textbf{H}\) outside the conductor drops to zero - because Maxwell's equations say that in regions free of currents (or charges, depending on the scenario), the magnetic fields change in characteristic ways.