Chapter 22 Magnetism
22.7 Magnetic Force on a Current-Carrying Conductor
Summary
- 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 vdvd is given by [latex]{F = qv_dB \;\text{sin} \;\theta}[/latex]. Taking [latex]{B}[/latex] to be uniform over a length of wire [latex]{l}[/latex] and zero elsewhere, the total magnetic force on the wire is then [latex]{F = (qv_dB \;\text{sin} \;\theta)(N)}[/latex], where [latex]{N}[/latex] is the number of charge carriers in the section of wire of length [latex]{l}[/latex]. Now, [latex]{N=nV}[/latex], where [latex]{n}[/latex] is the number of charge carriers per unit volume and [latex]{V}[/latex] is the volume of wire in the field. Noting that [latex]{V=Al}[/latex], where [latex]{A}[/latex] is the cross-sectional area of the wire, then the force on the wire is [latex]{F=(qv_dB \;\text{sin} \;\theta)(nAl)}[/latex]. Gathering terms,
Because [latex]{nqAv_d = I}[/latex] (see Chapter 20.1 Current),
is the equation for magnetic force on a length [latex]{l}[/latex] of wire carrying a current [latex]{I}[/latex] in a uniform magnetic field [latex]{B}[/latex], as shown in Figure 2. If we divide both sides of this expression by [latex]{l}[/latex], we find that the magnetic force per unit length of wire in a uniform field is [latex]{\frac{F}{l} = IB \;\text{sin} \;\theta}[/latex]. The direction of this force is given by RHR-1, with the thumb in the direction of the current [latex]{I}[/latex]. Then, with the fingers in the direction of [latex]{B}[/latex], a perpendicular to the palm points in the direction of [latex]{F}[/latex], as in Figure 2.
Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field
Calculate the force on the wire shown in Figure 1, given [latex]{B = 1.50 \;\text{T}}[/latex], [latex]{l = 5.00 \;\text{cm}}[/latex], and [latex]{I = 20.0 \;\text{A}}[/latex].
Strategy
The force can be found with the given information by using [latex]{F = IlB \;\text{sin} \;\theta}[/latex] and noting that the angle [latex]{\theta}[/latex] between [latex]{I}[/latex] and [latex]{B}[/latex] is [latex]{90 ^{\circ}}[/latex], so that [latex]{\text{sin} \;\theta = 1}[/latex].
Solution
Entering the given values into [latex]{F = IlB \;\text{sin} \theta}[/latex] yields
The units for tesla are [latex]{1 \;\text{T} = \frac{\text{N}}{\text{A} \cdot \; \text{m}}}[/latex]; 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 w