Question

A cylindrical wire of radius \(a\) is oriented with the \(z\) axis down its center line. The wire carries a nonuniform current down its length of density \(\mathbf{J}=b \rho \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}^{2}\) where \(b\) is a constant. ( \(a\) ) What total current flows in the wire? \((b)\) Find \(\mathbf{H}_{i n}(0<\rhoa)\), as a function of \(\rho ;(d)\) verify your results of parts \((b)\) and \((c)\) by using \(\nabla \times \mathbf{H}=\mathbf{J}\).

Short answer

In summary, we have calculated the total current flowing in the wire, which is given by \(I = \frac{2\pi b a^3}{3}\), and found the magnetic field inside and outside the wire. The magnetic field inside the wire (\(0 < \rho < a\)) is given by \(H_{in}(\rho) = \frac{2\pi b}{3}\rho^2\), while the magnetic field outside the wire (\(\rho > a\)) is given by \(H_{out}(\rho) = \frac{b a^3}{3\rho}\). We have verified that these results satisfy Ampère's law for both the inside and outside magnetic fields.

Step-by-step solution

Find the total current flowing in the wire

To find the total current flowing in the wire, we will compute the current density, sum it up over the area of the wire and integrate. Consider an infinitesimal cylindrical strip of radius \(\rho\) and thickness \(d\rho\). The area of the strip is given by \(dA = 2\pi\rho d\rho\). The current density in the strip is given by \(\mathbf{J} = b \rho \mathbf{a}_{z}\) where b is a constant. Now, let's calculate the total current I in the wire. To do this, we will sum up the current in every infinitesimal cylindrical strip: \[I = \int_0^a \mathbf{J} \cdot dA = \int_0^a b \rho \mathbf{a}_{z} \cdot (2\pi \rho d\rho \mathbf{a}_z)\]

Calculate the total current in the wire

Now, let's integrate the expression to find the total current I: \[I = 2 \pi b \int_0^a \rho^2 d\rho = 2\pi b \cdot \frac{a^3}{3}\] So, the total current flowing in the wire is given by \(I = \frac{2\pi b a^3}{3}\).

Find the magnetic field inside the wire (\(0 < \rho < a\))

We will use Ampère's law to find the magnetic field inside the wire. Ampère's law states: \[\oint_C \mathbf{H} \cdot d\mathbf{l} = I_{encl}\] We choose an Amperian loop that is a circle centered on the z-axis with radius \(\rho\) and infinitesimally small thickness \(d\rho\). Then the enclosed current \[I_{encl} = \int_0^\rho \mathbf{J} \cdot dA = 2 \pi b \int_0^\rho \rho'^2 d\rho' = 2\pi b \frac{\rho^3}{3}\] For the left side of Ampère's law, we have \(\oint_C \mathbf{H} \cdot d\mathbf{l} = H_{in} \oint_C dl = H_{in} (2\pi \rho)\). Thus: \[H_{in} (2\pi \rho) = 2\pi b \frac{\rho^3}{3}\] Now, let's find the expression for \(H_{in}\): \[H_{in}(\rho) = \frac{2\pi b}{3}\rho^2\]

Find the magnetic field outside the wire (\(\rho > a\))

For the magnetic field outside the wire, we will also use Ampère's law. In this case, we choose an Amperian loop that is a circle centered on the z-axis with radius \(\rho > a\). It is important to notice that, since the radius of the loop is larger than the wire radius, the whole wire carries a current enclosed by the loop. Therefore: \[I_{encl} = \frac{2\pi b a^3}{3}\] For the left side of Ampère's law, we have \(\oint_C \mathbf{H} \cdot d\mathbf{l} = H_{out} \oint_C dl = H_{out} (2\pi \rho)\). Thus, we have: \[H_{out} (2\pi \rho) = \frac{2\pi b a^3}{3}\] Now, let's find the expression for \(H_{out}\): \[H_{out}(\rho) = \frac{b a^3}{3\rho}\]

Verify the results of steps 3 and 4 using Ampère's law

We will now differentiate \(H_{in}\) and \(H_{out}\) with respect to \(\rho\) and check if the results satisfy Ampère's law: \(\nabla \times \mathbf{H} = \mathbf{J}\). For the magnetic field inside the wire, taking the curl of \(H_{in}\mathbf{a}_{\phi}\) (in cylindrical coordinates) gives: \[\nabla \times \mathbf{H}_{in} = \nabla \times \left(\frac{2\pi b}{3}\rho^2 \mathbf{a}_{\phi}\right) = \left(\frac{2\pi b}{\rho}\right) \mathbf{a}_z\] That verifies \(\nabla \times \mathbf{H}_{in} = \mathbf{J} = b \rho \mathbf{a}_{z}\) For the magnetic field outside the wire, we take the curl of \(H_{out}\mathbf{a}_{\phi}\) (in cylindrical coordinates): \[\nabla \times \mathbf{H}_{out} = \nabla \times \left(\frac{b a^3}{3\rho}\mathbf{a}_{\phi} \right) = 0\] This shows that there is no current outside the wire, which verifies our result in Step 4. Thus, the exercise is completed, and the results found satisfy Ampère's law.

Key concepts

Ampère's Law

Understanding Ampère's Law is fundamental when studying the magnetic effects of electric currents. It relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Mathematically, Ampère's Law is represented as \[ \textstyle \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{\text{encl}} \], where \( \mathbf{H} \) denotes the magnetic field, \( d\mathbf{l} \) is the infinitesimal element of the loop \( C \) and \( I_{\text{encl}} \) is the current enclosed by the loop.

This law is pivotal in determining the magnetic field inside and outside cylindrical wires. For a uniform current density, the magnetic field would be the same at equivalent distances from the center of the wire, due to the symmetry provided by the cylindrical geometry. In the exercises provided, Ampère’s Law helps find the magnetic fields both inside and outside the wire by using carefully chosen Amperian loops and considering the symmetry of the situation.

Current Density

Current density, represented by the symbol \( \mathbf{J} \), is a measure of the flow of electric charge per unit area, usually in the unit amperes per square meter (A/m²). In the problem at hand, the current density is not constant but varies with the radial distance \( \rho \) from the center of the wire, given as \( \mathbf{J} = b \rho \mathbf{a}_{z} \), where \( b \) is a constant. This non-uniformity is an essential factor influencing the calculation of the magnetic field within the wire.

A crucial step in the exercise is to calculate the total current flowing through the wire, which involves integrating the current density over the cross-sectional area of the cylindrical wire. Understanding current density and its integration in cylindrical coordinates is necessary to arrive at the correct expression for the total current.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that provides a natural way to describe points in space with cylindrical symmetry. In this system, a point \( P \) is represented by three values: \( \rho \) (the radial distance from the origin to the point), \( \phi \) (the angle between the x-axis and the line from the origin to the point's projection in the xy-plane), and \( z \) (the distance from the point to the xy-plane).

Using cylindrical coordinates simplifies calculations involving objects with circular cross-sections, such as wires and cylinders. The steps highlight their application in calculating the magnetic field generated by a cylindrical wire, where the radial symmetry of the problem suggests the use of \( \rho \) to express the currents and fields rather than Cartesian coordinates.

Magnetic Field of Cylindrical Wire

In the exercise, the magnetic field of a cylindrical wire, which carries a nonuniform current, is explored both inside and outside the wire. The magnetic field inside the wire (\( \mathbf{H}_{\text{in}} \)) depends on the distance from the center, while outside (\( \mathbf{H}_{\text{out}} \)), it varies inversely with the distance.

The use of Ampère's Law allows for the calculation of the field based on the enclosed current for any given radius. Inside the wire, the enclosed current increases with the radius up to the wire's outer boundary. Outside the wire, the enclosed current remains constant as it is defined by the wire's total current, leading to a decreasing magnetic field with increasing radius. Understanding these concepts shows why the exercise separates the calculations into two regions, each requiring careful consideration of the geometry and current distribution.