PHYS 208 Lecture 18

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Ampere's Law

Cylinder

Uniform current density through long cylindrical conductor: $\begin{array}{rl}\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }=B(r)\cdot (2\pi r)& ={\mu }_0{I}_{en}=\{\begin{array}{ll}{\mu }_{0}I\frac{r^2}{R^2}& r<R\\ {\mu }_{0}I& r>R\end{array}\\ B(r)& =\{\begin{array}{ll}\frac{{\mu }_{0}Ir}{2\pi R}& r<R\\ \frac{{\mu }_{0}I}{2\pi R}& r>R\end{array}\end{array}$

Plane of parallel wires

Draw a rectangular Amperian loop around a section of the sheet

$\begin{array}{rl}\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }={\oint }_1+{\oint }_2+{\oint }_3+{\oint }_4& ={\mu }_0{I}_{en}\\ BL+0+BL+0& ={\mu }_{0}NI\\ 2BL& ={\mu }_{0}nLI\\ B& =\frac{{\mu }_{0}nI}{2}\end{array}$

Toroid

Draw circular amperian loop around inside of torus.

$\begin{array}{rl}\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }=B(r)(2\pi r)& ={\mu }_0{I}_{en}={\mu }_{0}NI\\ B(r)& =\frac{{\mu }_{0}NI}{2\pi r}\end{array}$

Electromagnetic Induction

Creating a transformer around an iron core:

• Magnetic field created by coil connected to battery
• Pulse of current generated in coil on opposite side
• Amount of current depends on strength of magnetic field and the change in magnetic field

In general, it doesn't matter whether the magnet or the coil moves

Flux of Magnetic Field

${\mathrm{\Phi }}_B=\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{A}$

• Recall that this equals 0 for any closed surface
• For an open surface like the cross-sectional area of a wire loop, this value can be orbitrary.

Faraday's Law

For induced emf:

$\begin{array}{rl}{ℰ}& =-\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{\Phi }}_B\\ & =-\frac{\mathrm{d}}{\mathrm{d}t}(\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell })\end{array}$

The induced current produces a magnetic field that tries to counter the change in magnetic field (hence the negative sign).

Example

Suppose we take a rectangular loop (length $L$; width $W$) and insert it width first with a velocity $v$ into a uniform magnetic field $B$

First find the Flux: $\mathrm{\Phi }=BL(t)W=BvtW$

Calculate emf by Faraday's law:

${ℰ}=-BvW$

AC Generator

A loop spinning with angular velocity ω on an axis inside a uniform magnetic field.

$\begin{array}{rl}\mathrm{\Phi }(t)& =BA\cos \theta (t)\\ & =BA\cos \omega t\\ {ℰ}& =-\frac{\mathrm{d}}{\mathrm{d}t}(BA\cos (\omega t))\\ & =BA\omega \sin (\omega t)\end{array}$

Motional emf: Induced Electric Field

Moving a conductor through a uniform magnetic field:

• Magnetic force on charges inside conductor: $q\overrightarrow{v}\times \overrightarrow{B}$
• Charges move to opposite ends of the conductor, producing an increasing magnetic field
• In equlilibrium, electric force is same as magnetic force: $q\overrightarrow{v}\times \overrightarrow{B}=q\overrightarrow{E}$, therefore $\overrightarrow{v}\times \overrightarrow{B}=\overrightarrow{E}$