Question

A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d\Phi_B\) through a narrow strip of length \(l\) parallel to the axis, of width \(dr\), at a distance \(r\) from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(l\) of the cable is $$L = l \frac{\mu_0}{2\pi} \mathrm{ln}(\frac{b}{a})$$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length l of the cable.

Short answer

Magnetic field between conductors is \( B = \frac{\mu_0 i}{2\pi r}\). Total flux is \(\Phi_B = \frac{\mu_0 i \, l}{2\pi} \, \ln \left( \frac{b}{a} \right)\). Inductance \(L = \frac{\mu_0 l}{2\pi} \ln \left( \frac{b}{a} \right)\). Energy stored is \(U = \frac{\mu_0 l i^2}{4\pi} \ln \left( \frac{b}{a} \right)\).

Step-by-step solution

Apply Ampere's Law

To find the magnetic field between the conductors, we use Ampere's Law, which states \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \). For a path around the center conductor at a distance \(r\), the enclosed current \(I_{enc} = i\), and we get \(B \, (2\pi r) = \mu_0 i\), leading to \( B = \frac{\mu_0 i}{2\pi r} \).

Express Magnetic Flux Through a Strip

Consider a small area, a narrow strip of length \(l\) and width \(dr\), at a distance \(r\) from the axis. The area is \(dA = l \, dr\). The magnetic flux through this area is \(d\Phi_B = B \cdot dA \), hence \(d\Phi_B = \left( \frac{\mu_0 i}{2\pi r} \right) \, l \, dr \).

Integrate for Total Flux

To find the total magnetic flux \(\Phi_B\) between the conductors, integrate the expression for flux from \(a\) to \(b\): \(\Phi_B = \int_a^b \frac{\mu_0 i}{2\pi r} \cdot l \, dr = \frac{\mu_0 i \, l}{2\pi} \int_a^b \frac{1}{r} \, dr \). This evaluates to \(\Phi_B = \frac{\mu_0 i \, l}{2\pi} \, \ln \left( \frac{b}{a} \right)\).

Determine Inductance

Inductance \(L\) is defined by \( \Phi_B = L i \). Thus, equating \(\Phi_B = \frac{\mu_0 i \, l}{2\pi} \, \ln \left( \frac{b}{a} \right)\) to \(L i\), we solve for \(L\) to get \(L = \frac{\mu_0 l}{2\pi} \, \ln \left( \frac{b}{a} \right)\).

Calculate Magnetic Energy

Using the formula for energy \(U = \frac{1}{2} L i^2\), substituting the expression for \(L\) from the previous step, we find \(U = \frac{1}{2} \left( \frac{\mu_0 l}{2\pi} \ln \left( \frac{b}{a} \right) \right) i^2 = \frac{\mu_0 l i^2}{4\pi} \ln \left( \frac{b}{a} \right)\).

Key concepts

Magnetic Field Calculation

The concept of magnetic field calculation between conductors is a fundamental application of Ampere's Law. Ampere's Law helps in understanding how magnetic fields are generated by electric currents. In this scenario, we have two coaxial conductors with currents running in opposite directions, creating a magnetic field in the volume between them. To find the magnetic field at any point between the conductors, we use Ampere's Law, given by:\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \]Here, the integration path is a circular loop of radius \( r \) around the central conductor. The enclosed current \( I_{enc} \) is simply \( i \) since the inner conductor carries current \( i \). By symmetry, the magnetic field \( B \) is tangent to the circle, leading to:\[ B \times 2\pi r = \mu_0 i \]Solving for \( B \), we get:\[ B = \frac{\mu_0 i}{2\pi r} \]This equation provides the magnetic field at a distance \( r \) from the central conductor, allowing us to understand the field's behavior in the space between the conductors.

Inductance of a Coaxial Cable

The inductance of a coaxial cable is a measure of its ability to store magnetic energy when an electric current flows through it. Inductance depends on the magnetic flux linkage between the conductors. For our coaxial cable, we derive the inductance by first determining the total magnetic flux through an area perpendicular to the current flow. We use the formula:\[ \Phi_B = \int_a^b \frac{\mu_0 i}{2\pi r} \times l \times dr \]This integral calculates the total magnetic flux \( \Phi_B \) between the radii \( a \) and \( b \). Evaluating this integral results in:\[ \Phi_B = \frac{\mu_0 i \, l}{2\pi} \, \ln \left( \frac{b}{a} \right) \]Inductance \( L \) is defined by the relationship \( \Phi_B = L i \). Solving for \( L \), we have:\[ L = \frac{\mu_0 l}{2\pi} \, \ln \left( \frac{b}{a} \right) \] This provides the cable's inductance over a length \( l \), illustrating how physical dimensions affect its capacity to store magnetic energy.

Magnetic Energy Storage

Magnetic energy storage in a coaxial cable is quantified by considering the energy stored in the magnetic field due to the inductance. When a current flows through the cable, it builds up a magnetic field, storing energy that can be calculated using the inductance. The energy stored \( U \) is given by:\[ U = \frac{1}{2} L i^2 \]Substituting the expression for \( L \) from our previous section, we find:\[ U = \frac{1}{2} \left( \frac{\mu_0 l}{2\pi} \ln \left( \frac{b}{a} \right) \right) i^2 \]Simplifying, we obtain the energy stored in the magnetic field of a coaxial cable with a current \( i \) along its length \( l \):\[ U = \frac{\mu_0 l i^2}{4\pi} \ln \left( \frac{b}{a} \right) \]This expression highlights how the cable’s dimensions and current affect the energy stored. Understanding this helps in designing cables for specific energy storage requirements.

Magnetic Flux Integration

Magnetic flux integration is a key concept in determining how magnetic fields interlink through a given area, particularly in contexts like coaxial cables. In our scenario, we calculate the magnetic flux using the magnetic field we obtained from Ampere's Law.We consider a narrow strip of length \( l \) and width \( dr \) at a distance \( r \) from the axis. The area of this strip is \( dA = l \, dr \), and the magnetic flux through it is:\[ d\Phi_B = B \, dA = \left( \frac{\mu_0 i}{2\pi r} \right) l \, dr \]To find the total magnetic flux \( \Phi_B \) between the two conductors, we integrate \( d\Phi_B \) from \( a \) to \( b \):\[ \Phi_B = \int_a^b \frac{\mu_0 i}{2\pi r} l \, dr = \frac{\mu_0 i \, l}{2\pi} \int_a^b \frac{1}{r} \, dr \]Solving this gives:\[ \Phi_B = \frac{\mu_0 i \, l}{2\pi} \ln \left( \frac{b}{a} \right) \]This integral takes into account how the gradual changes over the radius from \( a \) to \( b \) contribute to total magnetic flux, which is crucial in calculating the inductance and stored energy.