# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

• Describe the effects of a magnetic force on a current-carrying conductor
• Calculate the magnetic force on a current-carrying conductor

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.C.3.1 The student is able to use right-hand rules to analyze a situation involving a current-carrying conductor and a moving electrically charged object to determine the direction of the magnetic force exerted on the charged object due to the magnetic field created by the current-carrying conductor. (S.P. 1.4)

Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.

Figure 5.22 The magnetic field exerts a force on a current-carrying wire in a direction given by the RHR-1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire because typical currents consist of very large numbers of moving charges.

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 $F=qvdBsinθ.F=qvdBsinθ.$ 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, $N=nV,N=nV,size 12{N= ital "nV"} {}$ 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 $V=Al,V=Al,size 12{V= ital "Al"} {}$ 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,

5.15 $F=(nqAvd)lBsinθ.F=(nqAvd)lBsinθ. size 12{F= $$ital "nqAv" rSub { size 8{d} }$$ ital "lB""sin"θ} {}$

Because $nqAvd=InqAvd=I size 12{ ital "nqAv" rSub { size 8{d} } =I} {}$ (see Current),

5.16 $F=IlBsinθF=IlBsinθ size 12{F= ital "IlB""sin"θ} {}$

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 Figure 5.23. If we divide both sides of this expression by $l,l,$ we find that the magnetic force per unit length of wire in a uniform field is $Fl=IBsinθ.Fl=IBsinθ.size 12{ { {F} over {l} } = ital "IB""sin"θ} {}$ The direction of this force is given by RHR-1, with the thumb in the direction of the current $I.I.size 12{I} {}$ Then, with the fingers in the direction of $B,B,size 12{B} {}$ a perpendicular to the palm points in the direction of $F,F,size 12{F} {}$ as in Figure 5.23.

Figure 5.23 The force on a current-carrying wire in a magnetic field is $F=IlBsinθ.F=IlBsinθ.size 12{F= ital "IlB""sin"θ} {}$ Its direction is given by RHR-1.

### Example 5.4Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field

Calculate the force on the wire shown in Figure 5.22, given $B=1.50 T,B=1.50 T,size 12{B=1 "." "50"" T"} {}$$l=5.00 cm,l=5.00 cm,size 12{l=5 "." "00"" cm"} {}$ and $I=20.0A.I=20.0A.size 12{I="20" "." 0 A} {}$

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 $90º,90º,$ so that $sinθ=1.sinθ=1.$

Solution

Entering the given values into $F=IlBsinθF=IlBsinθ size 12{F= ital "IlB""sin"θ} {}$ yields

5.17 $F=IlBsinθ=20.0 A0.0500 m1.50 T1 .F=IlBsinθ=20.0 A0.0500 m1.50 T1 . size 12{F= ital "IlB""sin"θ= left ("20" "." 0" A" right ) left (0 "." "0500"" m" right ) left (1 "." "50"" T" right ) left (1 right )} {}$

The units for tesla are $1 T=NA⋅m;1 T=NA⋅m; size 12{"1 T"= { {N} over {A cdot m} } } {}$ thus,

5.18 $F=1.50 N.F=1.50 N. size 12{F=1 "." "50"" N"} {}$

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 where magnetic force pumps fluids without moving mechanical parts. (See Figure 5.24.)

Figure 5.24 Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.

A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps. (Cell membranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (See Figure 5.25.) Existing MHD drives are heavy and inefficient—much development work is needed.

Figure 5.25 An MHD propulsion system in a nuclear submarine could produce significantly less turbulence than propellers and allow it to run more silently. The development of a silent drive submarine was dramatized in the book and the film The Hunt for Red October.