Chapter 22 Magnetism
22.7 Magnetic Force on a Current-Carrying Conductor
- 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 F=qvdBsinθF=qvdBsinθ. Taking BB size 12{B} {} to be uniform over a length of wire ll and zero elsewhere, the total magnetic force on the wire is then F=(qvdBsinθ)(N)F=(qvdBsinθ)(N) size 12{F= ( ital “qv” rSub { size 8{d} } B”sin”θ ) ( N ) } {}, where NN size 12{N} {} is the number of charge carriers in the section of wire of length ll size 12{l} {}. Now, N=nVN=nV size 12{N= ital “nV”} {}, where nn size 12{n} {} is the number of charge carriers per unit volume and VV size 12{V} {} is the volume of wire in the field. Noting that V=AlV=Al size 12{V= ital “Al”} {}, where AA size 12{A} {} is the cross-sectional area of the wire, then the force on the wire is F=(qvdBsinθ)(nAl)F=(qvdBsinθ)(nAl). Gathering terms,
Because nqAvd=InqAvd=I size 12{ ital “nqAv” rSub { size 8{d} } =I} {} (see Current),
is the equation for magnetic force on a length ll of wire carrying a current II in a uniform magnetic field BB, as shown in [link]. If we divide both sides of this expression by ll, we find that the magnetic force per unit length of wire in a uniform field is Fl=IBsinθFl=IBsinθ size 12{ { {F} over {l} } = ital “IB””sin”θ} {}. The direction of this force is given by RHR-1, with the thumb in the direction of the current II size 12{I} {}. Then, with the fingers in the direction of BB size 12{B} {}, a perpendicular to the palm points in the direction of FF size 12{F} {}, as in [link].
Calculate the force on the wire shown in [link], given B=1.50 TB=1.50 T size 12{B=1 “.” “50”” T”} {}, l=5.00 cml=5.00 cm size 12{l=5 “.” “00”” cm”} {}, and I=20.0AI=20.0A size 12{I=”20″ “.” 0 A} {}.
Strategy
The force can be found with the given information by using F=IlBsinθF=IlBsinθ size 12{F= ital “IlB””sin”θ} {} and noting that the angle θθ size 12{θ} {} between II size 12{I} {} and BB size 12{B} {} is 90º90º, so that sinθ=1sinθ=1.
Solution
Entering the given values into F=IlBsinθF=IlBsinθ size 12{F= ital “IlB””sin”θ} {} yields
The units for tesla are 1 T=NA⋅m1 T=NA⋅m size 12{“1 T”= { {N} over {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—they employ loops of wire and are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to a clever application wher