# Introduction To Magnetic Fields - MIT OpenCourseWare

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Chapter 8Introduction to Magnetic Fields8.1 Introduction. 18.2 The Definition of a Magnetic Field . 28.3 Magnetic Force on a Current-Carrying Wire. 3Example 8.1: Magnetic Force on a Semi-Circular Loop . 58.4 Torque on a Current Loop . 78.4.1 Magnetic force on a dipole . 10Animation 8.1: Torques on a Dipole in a Constant Magnetic Field. 118.5 Charged Particles in a Uniform Magnetic Field . 12Animation 8.2: Charged Particle Moving in a Uniform Magnetic Field. 148.6 Applications . 148.6.1 Velocity Selector. 158.6.2 Mass Spectrometer. 168.7 Summary. 178.8 Problem-Solving Tips . 188.9 Solved Problems . 198.9.18.9.28.9.38.9.4Rolling Rod. 19Suspended Conducting Rod. 20Charged Particles in Magnetic Field. 21Bar Magnet in Non-Uniform Magnetic Field . 228.10 Conceptual Questions . 238.11 Additional Problems . Force Exerted by a Magnetic Field. 23Magnetic Force on a Current Carrying Wire . 23Sliding Bar . 24Particle Trajectory. 25Particle Orbits in a Magnetic Field . 25Force and Torque on a Current Loop. 26Force on a Wire. 26Levitating Wire . 270

Introduction to Magnetic Fields8.1 IntroductionGWe have seen that a charged object produces an electric field E at all points in space. In aGsimilar manner, a bar magnet is a source of a magnetic field B . This can be readilydemonstrated by moving a compass near the magnet. The compass needle will line upalong the direction of the magnetic field produced by the magnet, as depicted in Figure8.1.1.Figure 8.1.1 Magnetic field produced by a bar magnetNotice that the bar magnet consists of two poles, which are designated as the north (N)and the south (S). Magnetic fields are strongest at the poles. The magnetic field linesleave from the north pole and enter the south pole. When holding two bar magnets closeto each other, the like poles will repel each other while the opposite poles attract (Figure8.1.2).Figure 8.1.2 Magnets attracting and repellingUnlike electric charges which can be isolated, the two magnetic poles always come in apair. When you break the bar magnet, two new bar magnets are obtained, each with anorth pole and a south pole (Figure 8.1.3). In other words, magnetic “monopoles” do notexist in isolation, although they are of theoretical interest.Figure 8.1.3 Magnetic monopoles do not exist in isolation8-2

GGHow do we define the magnetic field B ? In the case of an electric field E , we havealready seen that the field is defined as the force per unit charge:GG FeE q(8.1.1)GHowever, due to the absence of magnetic monopoles, B must be defined in a differentway.8.2 The Definition of a Magnetic FieldTo define the magnetic field at a point, consider a particle of charge q and moving at aGvelocity v . Experimentally we have the following observations:G(1) The magnitude of the magnetic force FB exerted on the charged particle is proportionalto both v and q.GGG(2) The magnitude and direction of FB depends on v and B .GGGG(3) The magnetic force FB vanishes when v is parallel to B . However, when v makes anGGGGangle θ with B , the direction of FB is perpendicular to the plane formed by v and B ,Gand the magnitude of FB is proportional to sin θ .(4) When the sign of the charge of the particle is switched from positive to negative (orvice versa), the direction of the magnetic force also reverses.Figure 8.2.1 The direction of the magnetic forceThe above observations can be summarized with the following equation:GG GFB q v B(8.2.1)2

The above expression can be taken as the working definition of the magnetic field at aGpoint in space. The magnitude of FB is given byFB q vB sin θ(8.2.2)The SI unit of magnetic field is the tesla (T):1 tesla 1 T 1NewtonNN 1 1(Coulomb)(meter/second)C m/sA mGAnother commonly used non-SI unit for B is the gauss (G), where 1T 104 G .GGGNote that FB is always perpendicular to v and B , and cannot change the particle’s speedv (and thus the kinetic energy). In other words, magnetic force cannot speed up or slowGdown a charged particle. Consequently, FB can do no work on the particle:GGG G GG G GdW FB d s q( v B) v dt q( v v) B dt 0(8.2.3)GThe direction of v , however, can be altered by the magnetic force, as we shall see below.8.3 Magnetic Force on a Current-Carrying WireWe have just seen that a charged particle moving through a magnetic field experiences aGmagnetic force FB . Since electric current consists of a collection of charged particles inmotion, when placed in a magnetic field, a current-carrying wire will also experience amagnetic force.Consider a long straight wire suspended in the region between the two magnetic poles.The magnetic field points out the page and is represented with dots ( ). It can be readilydemonstrated that when a downward current passes through, the wire is deflected to theleft. However, when the current is upward, the deflection is rightward, as shown in Figure8.3.1.Figure 8.3.1 Deflection of current-carrying wire by magnetic force3

To calculate the force exerted on the wire, consider a segment of wire of length A andcross-sectional area A, as shown in Figure 8.3.2. The magnetic field points into the page,and is represented with crosses ( X ).Figure 8.3.2 Magnetic force on a conducting wireGThe charges move at an average drift velocity v d . Since the total amount of charge in thissegment is Qtot q (nAA) , where n is the number of charges per unit volume, the totalmagnetic force on the segment isG GGG GG GFB Qtot v d B qnAA( v d B) I ( A B)(8.3.1)Gwhere I nqvd A , and A is a length vector with a magnitude A and directed along thedirection of the electric current.For a wire of arbitrary shape, the magnetic force can be obtained by summing over theforces acting on the small segments that make up the wire. Let the differential segment beGdenoted as d s (Figure 8.3.3).Figure 8.3.3 Current-carrying wire placed in a magnetic fieldThe magnetic force acting on the segment isGG Gd FB Id s B(8.3.2)GbG GFB I d s B(8.3.3)Thus, the total force isa4

where a and b represent the endpoints of the wire.GAs an example, consider a curved wire carrying a current I in a uniform magnetic field B ,as shown in Figure 8.3.4.Figure 8.3.4 A curved wire carrying a current I.Using Eq. (8.3.3), the magnetic force on the wire is given byGFB I( )baG GG Gd s B I A B(8.3.4)Gwhere A is the length vector directed from a to b. However, if the wire forms a closedloop of arbitrary shape (Figure 8.3.5), then the force on the loop becomesGFB IG( v d Gs ) B(8.3.5)Figure 8.3.5 A closed loop carrying a current I in a uniform magnetic field.GSince the set of differential length elements d s form a closed polygon, and their vectorGGGsum is zero, i.e., v d s 0 . The net magnetic force on a closed loop is FB 0 .Example 8.1: Magnetic Force on a Semi-Circular LoopConsider a closed semi-circular loop lying in the xy plane carrying a current I in thecounterclockwise direction, as shown in Figure 8.3.6.5

Figure 8.3.6 Semi-circular loop carrying a current IA uniform magnetic field pointing in the y direction is applied. Find the magnetic forceacting on the straight segment and the semicircular arc.Solution:GGGLet B Bˆj and F1 and F2 the forces acting on the straight segment and the semicircularparts, respectively. Using Eq. (8.3.3) and noting that the length of the straight segment is2R, the magnetic force isGF1 I (2 R ˆi ) ( B ˆj) 2 IRB kˆwhere k̂ is directed out of the page.GGTo evaluate F2 , we first note that the differential length element d s on the semicircle canGbe written as d s ds θˆ Rdθ ( sin θ ˆi cos θ ˆj) . The force acting on the length elementGd s isGG GdF2 Id s B IR dθ ( sin θ ˆi cos θ ˆj) ( B ˆj) IBR sin θ dθ kˆGHere we see that dF2 points into the page. Integrating over the entire semi-circular arc,we haveGπF2 IBR kˆ sin θ dθ 2 IBR kˆ0Thus, the net force acting on the semi-circular wire isGG G GFnet F1 F2 0This is consistent from our previous claim that the net magnetic force acting on a closedcurrent-carrying loop must be zero.6

8.4 Torque on a Current LoopWhat happens when we place a rectangular loop carrying a current I in the xy plane andGswitch on a uniform magnetic field B B ˆi which runs parallel to the plane of the loop,as shown in Figure 8.4.1(a)?Figure 8.4.1 (a) A rectangular current loop placed in a uniform magnetic field. (b) Themagnetic forces acting on sides 2 and 4.From Eq. 8.4.1, we see the magnetic forces acting on sides 1 and 3 vanish because theGGGlength vectors A1 b ˆi and A3 b ˆi are parallel and anti-parallel to B and their crossproducts vanish. On the other hand, the magnetic forces acting on segments 2 and 4 arenon-vanishing:G F2 I ( a ˆj) ( B ˆi ) IaB kˆ G F4 I (a ˆj) ( B ˆi ) IaB kˆ(8.4.1)GGwith F2 pointing out of the page and F4 into the page. Thus, the net force on therectangular loop isGG G G G GFnet F1 F2 F3 F4 0(8.4.2)GGas expected. Even though the net force on the loop vanishes, the forces F2 and F4 willproduce a torque which causes the loop to rotate about the y-axis (Figure 8.4.2). Thetorque with respect to the center of the loop isG b G b G b b τ ˆi F2 ˆi F4 ˆi IaB kˆ ˆi IaB kˆ 2 2 2 2 ( IabB IabB 2 2 ˆˆˆ j IabB j IAB j )()(8.4.3)7

where A ab represents the area of the loop and the positive sign indicates that therotation is clockwise about the y-axis. It is convenient to introduce the area vectorGA A nˆ where n̂ is a unit vector in the direction normal to the plane of the loop. Thedirection of the positive sense of n̂ is set by the conventional right-hand rule. In our case,we have n̂ kˆ . The above expression for torque can then be rewritten asG GGτ IA B(8.4.4)GNotice that the magnitude of the torque is at a maximum when B is parallel to the planeGof the loop (or perpendicular to A ).GConsider now the more general situation where the loop (or the area vector A ) makes anangle θ with respect to the magnetic field.Figure 8.4.2 Rotation of a rectangular current loopFrom Figure 8.4.2, the lever arms and can be expressed as:()G bGr2 sin θ ˆi cos θ kˆ r42(8.4.5)and the net torque becomesG GbG G G G Gτ r2 F2 r4 F4 2r2 F2 2 sin θ ˆi cos θ kˆ IaB kˆ2G G IabB sin θ ˆj IA B() ()(8.4.6)For a loop consisting of N turns, the magnitude of the toque isτ NIAB sin θ(8.4.7)GGThe quantity NIA is called the magnetic dipole moment µ :GGµ NI A(8.4.8)8

GFigure 8.4.3 Right-hand rule for determining the direction of µGGThe direction of µ is the same as the area vector A (perpendicular to the plane of theloop) and is determined by the right-hand rule (Figure 8.4.3). The SI unit for the magneticGdipole moment is ampere-meter2 (A m 2 ) . Using the expression for µ , the torque exertedon a current-carrying loop can be rewritten asG G Gτ µ B(8.4.9)G G GThe above equation is analogous to τ p E in Eq. (2.8.3), the torque exerted on anGGelectric dipole moment p in the presence of an electric field E . Recalling that theG Gpotential energy for an electric dipole is U p E [see Eq. (2.8.7)], a similar form isexpected for the magnetic case. The work done by an external agent to rotate themagnetic dipole from an angle θ0 to θ is given byθθθ0θ0Wext τ dθ ′ ( µ B sin θ ′)dθ ′ µ B ( cos θ 0 cos θ ) U U U 0(8.4.10)Once again, Wext W , where W is the work done by the magnetic field. ChoosingU 0 0 at θ 0 π / 2 , the dipole in the presence of an external field then has a potentialenergy ofG GU µ B cos θ µ B(8.4.11)GGThe configuration is at a stable equilibrium when µ is aligned parallel to B , making U aGGminimum with U min µ B . On the other hand, when µ and B are anti-parallel,U max µ B is a maximum and the system is unstable.9

8.4.1Magnetic force on a dipoleAs we have shown above, the force experienced by a current-carrying rectangular loop(i.e., a magnetic dipole) placed in a uniform magnetic field is zero. What happens if themagnetic field is non-uniform? In this case, there will be a net force acting on the dipole.GConsider the situation where a small dipole µ is placed along the symmetric axis of a barmagnet, as shown in Figure 8.4.4.Figure 8.4.4 A magnetic dipole near a bar magnet.The dipole experiences an attractive force by the bar magnet whose magnetic field is nonuniform in space. Thus, an external force must be applied to move the dipole to the right.The amount of force Fext exerted by an external agent to move the dipole by a distance x is given byFext x Wext U µ B( x x) µ B( x) µ[ B( x x) B ( x)](8.4.12)where we have used Eq. (8.4.11). For small x , the external force may be obtained asFext µ[ B ( x x) B ( x)]dB µ xdx(8.4.13)which is a positive quantity since dB / dx 0 , i.e., the magnetic field decreases withincreasing x. This is precisely the force needed to overcome the attractive force due to thebar magnet. Thus, we haveFB µdB d G G (µ B)dx dx(8.4.14)GMore generally, the magnetic force experienced by a dipole µ placed in a non-uniformGmagnetic field B can be written asGG GFB (µ B)(8.4.15)where10

ˆ ˆ ˆi j k x y z(8.4.16)is the gradient operator.Animation 8.1: Torques on a Dipole in a Constant Magnetic Field“ To understand this point, we have to consider that a [compass] needlevibrates by gathering upon itself, because of it magnetic condition andpolarity, a certain amount of the lines of force, which would otherwisetraverse the space about it ”Michael Faraday [1855]Consider a magnetic dipole in a constant background field. Historically, we note thatFaraday understood the oscillations of a compass needle in exactly the way we describehere. We show in Figure 8.4.5 a magnetic dipole in a “dip needle” oscillating in themagnetic field of the Earth, at a latitude approximately the same as that of Boston. Themagnetic field of the Earth is predominantly downward and northward at these Northernlatitudes, as the visualization indicates.Figure 8.4.5 A magnetic dipole in the form of a dip needle oscillates in the magneticfield of the Earth.To explain what is going on in this visualization, suppose that the magnetic dipole vectoris initially along the direction of the earth’s field and rotating clockwise. As the dipolerotates, the magnetic field lines are compressed and stretched. The tensions and pressuresassociated with this field line stretching and compression results in an electromagnetictorque on the dipole that slows its clockwise rotation. Eventually the dipole comes to rest.But the counterclockwise torque still exists, and the dipole then starts to rotatecounterclockwise, passing back through being parallel to the Earth’s field again (wherethe torque goes to zero), and overshooting.As the dipole continues to rotate counterclockwise, the magnetic field lines are nowcompressed and stretched in the opposite sense. The electromagnetic torque has reversedsign, now slowing the dipole in its counterclockwise rotation. Eventually the dipole willcome to rest, start rotating clockwise once more, and pass back through being parallel to11

the field, as in the beginning. If there is no damping in the system, this motion continuesindefinitely.Figure 8.4.6 A magnetic dipole in the form of a dip needle rotates oscillates in themagnetic field of the Earth. We show the currents that produce the earth’s field in thisvisualization.What about the conservation of angular momentum in this situation? Figure 8.4.6 showsa global picture of the field lines of the dip needle and the field lines of the Earth, whichare generated deep in the core of the Earth. If you examine the stresses transmittedbetween the Earth and the dip needle in this visualization, you can convince yourself thatany clockwise torque on the dip needle is accompanied by a counterclockwise torque onthe currents producing the earth’s magnetic field. Angular momentum is conserved bythe exchange of equal and opposite amounts of angular momentum between the compassand the currents in the Earth’s core.8.5 Charged Particles in a Uniform Magnetic FieldIf a particle of mass m moves in a circle of radius r at a constant speed v, what acts on theparticle is a radial force of magnitude F mv 2 / r that always points toward the centerand is perpendicular to the velocity of the particle.GIn Section 8.2, we have also shown that the magnetic force FB always points in theGGdirection perpendicular to the velocity v of the charged particle and the magnetic field B .GGSince FB can do not work, it can only change the direction of v but not its magnitude.GWhat would happen if a charged particle moves through a uniform magnetic field B withGGits initial velocity v at a right angle to B ? For simplicity, let the charge be q and theGGdirection of B be into the page. It turns out that FB will play the role of a centripetalforce and the charged particle will move in a circular path in a counterclockwise direction,as shown in Figure 8.5.1.12

GGFigure 8.5.1 Path of a charge particle moving in a uniform B field with velocity vGinitially perpendicular to B .WithqvB mv 2r(8.5.1)the radius of the circle is found to ber mvqB(8.5.2)The period T (time required for one complete revolution) is given byT 2π r 2π mv 2π m vv qBqB(8.5.3)Similarly, the angular speed (cyclotron frequency) ω of the particle can be obtained asω 2π f v qB r m(8.5.4)If the initial velocity of the charged particle has a component parallel to the magneticGfield B , instead of a circle, the resulting trajectory will be a helical path, as shown inFigure 8.5.2:Figure 8.5.2 Helical path of a charged particle in an external magnetic field. The veloci

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