Magnetic Field Created by a Long Straight Current-Carrying Wire: Right Hand Rule 2
Magnetic fields have both direction and magnitude. As noted before, one way to explore the direction of a magnetic field is with compasses, as shown for a long straight current-carrying wire in this figure. Hall probes can determine the magnitude of the field. The field around a long straight wire is found to be in circular loops. The right hand rule 2 (RHR-2) emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.
(a) Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. (b) Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.
The magnetic field strength (magnitude) produced by a long straight current-carrying wire is found by experiment to be
\(B=\cfrac{{\mu }_{0}I}{2\mathrm{\pi r}}\phantom{\rule{0.25em}{0ex}}(\text{long straight wire}),\)
where \(I\) is the current, \(r\) is the shortest distance to the wire, and the constant \({\mu }_{0}=4\pi \phantom{\rule{0.15em}{0ex}}×\phantom{\rule{0.15em}{0ex}}{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}T\cdot \text{m/A}\) is the permeability of free space. \(({\mu }_{0}\) is one of the basic constants in nature. We will see later that \({\mu }_{0}\) is related to the speed of light.) Since the wire is very long, the magnitude of the field depends only on distance from the wire \(r\), not on position along the wire.
Example: Calculating Current that Produces a Magnetic Field
Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth’s at a distance of 5.0 cm from the wire.
Strategy
The Earth’s field is about \(5\text{.}0×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}T\), and so here \(B\) due to the wire is taken to be \(1\text{.}0×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}T\). The equation \(B=\cfrac{{\mu }_{0}I}{2\mathrm{\pi r}}\) can be used to find \(I\), since all other quantities are known.
Solution
Solving for \(I\) and entering known values gives
\(\begin{array}{lll}I& =& \cfrac{2\pi \text{rB}}{{\mu }_{0}}=\cfrac{2\pi (5.0×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}m)(1.0×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}T)}{4\pi ×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}T\cdot \text{m/A}}\\ & =& \text{25 A.}\end{array}\)
Discussion
So a moderately large current produces a significant magnetic field at a distance of 5.0 cm from a long straight wire. Note that the answer is stated to only two digits, since the Earth’s field is specified to only two digits in this example.