# Magnetic field due to a long straight wire & force between two parallel wires #### Everything You Need in One Place

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##### Intros
###### Lessons
1. Magnetic field due to a long straight wire
2. Magnetic field midway between two currents
3. Forces between two parallel wires
##### Examples
###### Lessons
1. In which diagram would an external magnetic field $\overline{B}$, cause two current-carrying wires to move towards one another? 1. Which of the four diagrams below correctly depicts the magnetic field found on either side of a current-carrying wire? 1. A long pair of wires conduct 25.0A of current to and from an instrument. If the insulated wires are of negligible diameter but are 2.88mm apart, what is the magnetic field 10.00cm from their midpoint? ###### Topic Notes

In this lesson, we will learn:

• Magnetic field due to a long straight wire
• Magnetic field midway between two currents
• Forces between two parallel wires

Notes:

An electric current produces a magnetic field
• The magnetic field surrounding the electric current in a long straight wire is such that the field lines are circles with the wire at the center.
• The field strength at a given point would be greater if the current flowing in the wire were greater; $B \propto I$
• The filed strength would be less at points farther from the wire $B \propto \frac{1}{r}$;

• $\quad B \propto I$
$\quad B \propto \frac{1}{r} \qquad B \propto \frac{1}{r} \quad \Rightarrow \quad B =$ $\large \frac{\mu_{0}}{2 \pi} \frac{I}{r}$,

The value of the constant $\mu_{0}$, which is called the permeability of free space, is $\mu_{0}$ = 4$\pi$ × 10-7 $T.m/A$ Magnetic Field Midway Between Two Currents

Two parallel wires 10.0cm apart carry currents in opposite directions. Current $I_{1}$ = 5.0A is out of the page, $I_{2}$=7.0 A is into the page. Determine the magnitude and direction of the magnetic field halfway between the two wires. $B_{1} =$ $\large \frac{\mu_{0} I_{1}} {2\pi r} = \frac{(4\pi \, \times \, 10^{-7} \, T\, . \, m/A) (5. 0A) } {2 \pi (0.050m) }$ = 2.0 × 10-5 $T$

$B_{2} =$ $\large \frac{\mu_{0} I_{2}} {2\pi r} = \frac{(4\pi \, \times \, 10^{-7} \, T\, . \, m/A) (7. 0A) } {2 \pi (0.050m) }$ = 2.8 × 10-5 $T$

The total filed is up with the magnitude of

$B = B_{1} +B_{2} = 4.8 \times 10^{-5} T$

Forces Between Two Parallel Wires
• Consider two long parallel wires separated by a distance $d$. They carry currents $I_{1}$ and $I_{2}$, respectively. Each current produces a magnetic field that is felt by the other, so each must exert a force on the other.

• • Magnetic field $B_{1}$ produced by $I_{1}, \, B_{1} =$ $\large \frac{\mu_{0} I_{1}} {2 \pi d}$
• The force $F_{2}$ is exerted by $B_{1}$ on a length of $I_{2}$ of wire 2, carrying current $I_{2}$, $F_{2}$ = $I_{2}$ $B_{1}$ $I_{2}$
• Substitute $B_{1}$ into $F_{2}$ formula to get the final equation;

• $F_{2} =$ $\large \frac{\mu_{0}} {2 \pi} \frac{I_{1} I_{2}} {d} I_{2}$

• Parallel currents in the same direction exert an attractive force on each other
• Antiparallel currents (in opposite directions) exert a repulsive force on each other. 