Question

A long copper wire carries a current of $I$ ampere. Calculate the magnetic flux per meter of the wire for a plane surface $S$ inside the wire as shown in figure.

Answer 1

The magnetic flux per meter of the wire for the plane surface $S$ is $\frac{\mu_0 I R \Delta \phi}{6\pi}$, where $\Delta \phi$ is the angle of the sector in radians.

Answer 2

1. Magnetic Field Inside the Wire: Using Ampere's Law, the magnetic field at a radial distance $r$ from the center of a long wire of radius $R$ carrying current $I$ is given by $B(r) = \frac{\mu_0 I r}{2\pi R^2}$ for $r < R$. The magnetic field lines are circles concentric with the wire's axis.

2. Surface Geometry: The surface $S$ is a planar sector originating from the center $O$ and extending to the radius $R$. Let the angle subtended by the sector at the center be $\Delta \phi$. The question asks for the magnetic flux per meter of the wire, so we consider a length $L=1$ meter along the wire. Thus, the surface $S$ is a planar sector with radial extent $0$ to $R$ and angle $\Delta \phi$, and length $L=1$ along the wire.

3. Flux Calculation: The magnetic flux through the surface $S$ for a length $L=1$ meter is given by: $\Phi = \int_S \vec{B} \cdot d\vec{A}$ The magnetic field $\vec{B}$ at a distance $r$ is tangential to the circle of radius $r$. For a planar sector surface $S$, the magnetic field lines are circulating. To calculate the flux, we consider the component of the magnetic field perpendicular to the surface. Let's consider a small area element $dA$ on the surface $S$ at a radial distance $r$. The magnetic field at $r$ is $B(r) = \frac{\mu_0 I r}{2\pi R^2}$. The flux per meter of wire is given by: $\frac{\Phi}{L} = \int_0^R \int_0^{\Delta \phi} B(r) \cdot r dr d\theta$ This integral represents the flux through a surface where the magnetic field is perpendicular to the area element. Given the tangential nature of the magnetic field and the planar nature of the surface $S$, we need to consider the component of $B$ that contributes to the flux. The correct approach is to consider the flux through the planar sector. The magnetic field lines are circulating. For a planar sector, the flux is obtained by integrating the magnetic field component that is perpendicular to the plane of the sector. If the surface $S$ is a planar sector, the magnetic field lines are tangential to the circles. The flux is calculated by considering the magnetic field's contribution across the area. The flux through the surface $S$ per meter of wire is given by integrating the magnetic field component perpendicular to the surface. $\frac{\Phi}{L} = \int_0^R \frac{\mu_0 I r}{2\pi R^2} \cdot r dr \cdot \Delta \phi$ $\frac{\Phi}{L} = \frac{\mu_0 I \Delta \phi}{2\pi R^2} \int_0^R r^2 dr$ $\frac{\Phi}{L} = \frac{\mu_0 I \Delta \phi}{2\pi R^2} \left[ \frac{r^3}{3} \right]_0^R$ $\frac{\Phi}{L} = \frac{\mu_0 I \Delta \phi}{2\pi R^2} \cdot \frac{R^3}{3}$ $\frac{\Phi}{L} = \frac{\mu_0 I R \Delta \phi}{6\pi}$ The magnetic flux per meter of the wire for the plane surface $S$ is $\frac{\mu_0 I R \Delta \phi}{6\pi}$, where $\Delta \phi$ is the angle of the sector in radians.