122 Magnetic Force on a Current-Carrying Conductor
122 Magnetic Force on a Current-Carrying Conductor
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Learning Objectives
- Describe the effects of a magnetic force on a current-carrying conductor.
- Calculate the magnetic force on a current-carrying conductor.
Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.
We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity \({v}_{d}\) is given by \(F={\text{qv}}_{d}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \). Taking \(B\) to be uniform over a length of wire \(l\) and zero elsewhere, the total magnetic force on the wire is then \(F=\left({\text{qv}}_{d}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)\left(N\right)\), where \(N\) is the number of charge carriers in the section of wire of length \(l\). Now, \(N=\text{nV}\), where \(n\) is the number of charge carriers per unit volume and \(V\) is the volume of wire in the field. Noting that \(V=\text{Al}\), where \(A\) is the cross-sectional area of the wire, then the force on the wire is \(F=\left({\text{qv}}_{d}B\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)\left(\text{nAl}\right)\). Gathering terms,
Because \({\text{nqAv}}_{d}=I\) (see Current),
is the equation for magnetic force on a length \(l\) of wire carrying a current \(I\) in a uniform magnetic field \(B\), as shown in (Figure). If we divide both sides of this expression by \(l\), we find that the magnetic force per unit length of wire in a uniform field is \(\frac{F}{l}=\text{IB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \). The direction of this force is given by RHR-1, with the thumb in the direction of the current \(I\). Then, with the fingers in the direction of \(B\), a perpendicular to the palm points in the direction of \(F\), as in (Figure).
Calculate the force on the wire shown in (Figure), given \(B=1\text{.}\text{50 T}\), \(l=5\text{.}\text{00 cm}\), and \(I=\text{20}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{A}\).
Strategy
The force can be found with the given information by using \(F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) and noting that the angle \(\theta \) between \(I\) and \(B\) is \(\text{90º}\), so that \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1\).
Solution
Entering the given values into \(F=\text{IlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) yields
The units for tesla are \(\text{1 T}=\frac{N}{A\cdot m}\); thus,
Discussion
This large magnetic field creates a significant force on a small length of wire.
Magnetic force on current-carrying conductors is used to convert electric energy to work. (Motors are a prime example—th