8.02 Spring 2003VIII. Magnetic Fields - Worked ExamplesExample 5: Bar magnet in non-uniform magnetic fieldMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2003 VIII. Magnetic Fields - Worked Examples Example 1: Rolling rod A rod with a mass m and a radius R is mounted on two parallel rails of length L separated by a distance d, as shown in the figure below. The rod carries a current I and rolls without slipping along the rails which are placed in a uniform magnetic field G(directions shown in the figure). If the rod is initially at rest, what is its speed as it leaves the rails? B Solution: Using the coordinate system shown on the right, the magnetic force acting on the rod is given by ˆˆˆ()( )BIId B IdB=×= ×− =FdB i kjGGG (1.1) The total work done by the magnetic force on the rod as it moves through the region is (1.2)BBWdFLId=⋅ =∫Fs=GGBL By the work-energy theorem, W must be equal to the change in kinetic energy: 021122Kmv I2ω∆= + (1.3) where both translation and rolling are involved. Since the moment of inertia of the rod is given by , and the condition of rolling with slipping implies 2/2ImR= /vRω= , we have 222211 11 3222 244mR v22IdBL mv mv mv mvR=+ =+= (1.4) Thus, the speed of the rod as it leaves the rails is 43IdBLvm= (1.5) Example 2: Magnetic dipole moment in B field A current loop with magnetic dipole moment µG is placed in a uniform magnetic field BG. Show that its potential energy is given by U=−⋅µBGG (2.1) Solution: The magnetic field exerts a torque =×τµBGGG of magnitude sinBτµθ= on the dipole, tending to turn the dipole moment in the direction of decreasing θ. We can choose the potential energy U to be zero when the dipole moment is at an angle /2θπ= to the magnetic field. Thus, the energy is given by []/2/20sin cos cUBdB Bθθππosµθθ µ θ µ θ−= = − =−∫ (2.2) or we can write U=−⋅µBGG (2.3) The potential energy U represents the amount of work which needs to be done by an external agent to orient the dipole µGat an angle θ with the magnetic field G. B 1Example 3: Suspended conductor Suppose a conductor having a mass density λ kg/m is suspended by two flexible wires in a uniform magnetic field Gwhich points into the page (see figure below). inB If the tension on the wires is zero, what are the magnitude and the direction of the current in the conductor? Solution: In order that the tension in the wires be zero, the magnetic force GG acting on the conductor must exactly cancel the downward gravitational force BI=×FLBGˆgmg=−FkG. For to point in the +z direction, we must have BFGˆLL=jG, i.e., the current flows to the right, so that ˆˆ ˆˆˆ()( ) ( )BIIL B ILB ILB=×= ×− =− ×=+FLBjijiGGGk (3.1) The magnitude of the current can be obtain from ILB mg= (3.2) or mg gIBLBλ== (3.3) 2Example 4: Charged particles in B field Particle A with charge q and mass mA and particle B with charge 2q and mass mB, are accelerated from rest by a potential difference V , and subsequently deflected by a uniform magnetic field into semicircular paths. The radii of the particle A and B are R and 2R, respectively. The direction of the magnetic field is perpendicular to the velocity of the particle. What is their mass ratio? Solution: The kinetic energy gained by the charges is equal to 212mv qV= (4.1) which yields 2qVvm= (4.2) The charges move in semicircle, since the magnetic force points radially inward and provides the source of the centripetal force: 2mvqvBr= (4.3) The radius of the circle can be readily obtained as: 212mv m qV mVrqB qB m B q== = (4.4) which shows that r is proportional to . The mass ratio can then be obtained from 1/ 2(/)mq 1/ 2 1/ 21/ 2 1/2(/) (/) (/) 2 (/2)AAA ABBB Brmq mqRrmq Rmq=⇒= (4.5) which gives 18ABmm= (4.6) 3Example 5: Bar magnet in non-uniform magnetic field A bar magnet with its north pole up is placed along the symmetric axis below a horizontal conducting ring carrying current I, as shown in the figure below. What is the force on the ring? Solution: The magnetic force acting on a small differential current-carrying element Id sGon the ring is given by dI, where d=×FsGGGB BGis the magnetic field due to the bar magnet. Due to the axial symmetry, the magnetic field lines will intersect the loop at right angles (figure below), so that the x component of the force will exactly cancel. Thus, net force experienced by Id sG is sin ( )sinydF dF IdsBθθ== (5.1) The total force acting on the ring then becomes sin 2 sinyyFdFIB dsrIBθπ== =∫∫vvθ (5.2) The force points in the +y direction and therefore is repulsive.
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