Question

A wire of radius \(R\) carries a current \(i\). The current density is given by \(J=J_{0}(1-r / R),\) where \(r\) is measured from the center of the wire and \(J_{0}\) is a constant. Use Ampere's Law to find the magnetic field inside the wire at a distance \(r

Short answer

Question: Determine the magnetic field inside a wire of radius R carrying a current i with a current density given by \(J=J_{0}(1-r/R)\). Answer: The magnetic field inside the wire at a distance \(r<R\) from the central axis is given by the expression: $$B(r) = \frac{\mu_0 J_{0} R^2}{2} \left( 1 - \frac{r^2}{R^2} \right)$$

Step-by-step solution

Calculate the current through the wire

To find the total current through the wire, we need to integrate the current density \(J\) over the entire cross-sectional area of the wire. The current density is given by \(J=J_{0}(1-r/R)\), where r is the radial distance from the center of the wire. We can express the current through a small area element \(dA\) at distance r as \(dI = J dA = J_{0}(1-r/R) dA\). The area element in polar coordinates is given by \(dA = r dr d\theta\). Therefore, the current through this element is \(dI = J_{0}(1-r/R) r dr d\theta\). Now, we need to integrate dI over the entire cross-sectional area of the wire to find the total current i. We will utilize the limits of integration \(0<r<R\) and \(0<\theta<2\pi\). $$i = \int_{0}^{R} \int_{0}^{2\pi} J_{0}(1-r/R) r dr d\theta$$

Apply Ampere's Law

Ampere's Law states that the circulation of the magnetic field \(B\) around a closed path is equal to the permeability constant \(\mu_0\) times the current through the path: $$\oint B dl = \mu_0 i$$ To apply Ampere's Law, we will consider a circular path of radius \(r<R\) around the central axis of the wire. The magnetic field B inside the wire will be tangential to the circular path and have equal magnitude at each point around the path. Let \(dl\) be a small length element along the path. Then the integral of B around the circular path will be: $$\oint B dl = B \oint dl = B (2\pi r)$$ Calculating the current through a path of radius r<R, we will have: $$i_r = \int_{0}^{r} \int_{0}^{2\pi} J_{0}(1-r'/R) r' dr' d\theta = 2\pi J_{0} \int_{0}^{r}(1-r'/R) r' dr'$$ By substituting the expression for \(i_r\) into Ampere's Law, we have: $$B(2\pi r) = \mu_0 i_r$$ Now, we will find the magnetic field B.

Simplify the expression for the magnetic field

Equating the expressions for the magnetic field from Ampere's Law, we have: $$B(2\pi r) = \mu_0 (2\pi J_{0} \int_{0}^{r}(1-r'/R) r' dr')$$ Simplifying the expression for the magnetic field, we get: $$B(r) = \frac{\mu_0 J_{0}}{2\pi r} \int_{0}^{r}(1-r'/R) r' dr'$$ Integrating and simplifying further, we obtain: $$B(r) = \frac{\mu_0 J_{0} R^2}{2} \left( 1 - \frac{r^2}{R^2} \right)$$ This is the expression for the magnetic field inside the wire at a distance \(r<R\) from the central axis.

Key concepts

Current Density

Current density is a measure of how much electric current flows through a specific area of a conductor. It is often denoted by the symbol 'J' and is calculated as the current 'I' divided by the cross-sectional area 'A' through which it flows, represented by the formula:
\( J = \frac{I}{A} \).
In the context of our exercise, current density varies with the distance from the center of a wire; it's not uniform across the cross-section. The formula given, \(J = J_0(1 - \frac{r}{R})\), indicates that the current density decreases linearly from its maximum value at the center to zero at the surface, with \(J_0\) being the maximum current density at the center and 'r' representing the radial distance from the center. This varying current density is integral to determining the internal magnetic field using Ampere's Law.

Magnetic Field Inside Conductor

Inside a conductor, the magnetic field is generated by the movement of the electric charges or the flow of current. According to Ampere's Law, this magnetic field can be found around a current carrying conductor. Ampere's Law states that for a closed loop, the sum of the magnetic field times the length of the loop \(\oint B \cdot dl\) is proportional to the current passing through any surface bounded by the loop.
For our exercise, to find the magnetic field inside the wire, we consider a circular path with radius 'r' less than the radius of the wire. The magnetic field is tangential to this path and is constant in magnitude along it. By applying Ampere's Law, we relate the magnetic field to the current enclosed by the path. Through integration, we obtain a formula for the magnetic field based on the varying current density inside the conductor.

Integration in Physics

Integration is a fundamental mathematical tool in physics, especially in the case involving continuous variables. It allows for the calculation of quantities when a distribution is variable across a certain dimension—like current density over a wire's cross-section. In our exercise, integration is employed to add up the infinitesimally small contributions of current (through small elements of area 'dA') over the entire cross-section to find out the total current 'i'.
The integral symbol \(\int\) is the key operator, and the calculations can range from simple to complex, depending on the function to be integrated. The limits of the integral must accurately reflect the physical dimensions of the problem, such as integrating the current density from the center of the wire \(r = 0\) to any point within the wire \(r < R\). The solution demonstrates the process of integrating the current density to find the enclosed current, which is then used in Ampere's Law to calculate the magnetic field inside the conductor.