VIII. Magnetic Fields - Worked Examples

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 003 VIII. Magnetic Fields - Worked Examples Example : 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 (directions shown in the figure). If the rod is initially at rest, what is its speed as it leaves the rails? Using the coordinate system shown on the right, the magnetic force acting on the rod is given by F ( ˆ) ( ˆ ) ˆ = Id = Idi k = Idj (.) The total work done by the magnetic force on the rod as it moves through the region is W = d F L= IdL F s = (.) y the work-energy theorem, W must be equal to the change in kinetic energy: 0

2 K = mv + Iω (.3) where both translation and rolling are involved. Since the moment of inertia of the rod is given by I = mr /, and the condition of rolling with slipping implies ω = v/ R, we have mr v 3 IdL = mv + = mv + mv = mv R 4 4 (.4) Thus, the speed of the rod as it leaves the rails is 4IdL v = (.5) 3m Example : Magnetic dipole moment in field A current loop with magnetic dipole moment µ is placed in a uniform magnetic field. Show that its potential energy is given by U = µ (.) The magnetic field exerts a torque τ = µ of magnitude τ = µ sin on the dipole, tending to turn the dipole moment in the direction of decreasing. We can choose the potential energy U to be zero when the dipole moment is at an angle = π / to the magnetic field. Thus, the energy is given by or we can write π / [ ] / U 0= µ sin d = µ cos = µ cos (.) π U = µ (.3) The potential energy U represents the amount of work which needs to be done by an external agent to orient the dipole µ at an angle with the magnetic field.

3 Example 3: Suspended conductor Suppose a conductor having a mass density λ kg/m is suspended by two flexible wires in a uniform magnetic field which points into the page (see figure below). in If the tension on the wires is zero, what are the magnitude and the direction of the current in the conductor? In order that the tension in the wires be zero, the magnetic force F = IL acting on the conductor must exactly cancel the downward gravitational force F = mgkˆ. g For F to point in the +z direction, we must have right, so that L= Lˆj, i.e., the current flows to the F ( ˆ) ( ˆ) ( ˆ ˆ) ˆ = IL = I Lj i = IL j i =+ ILk (3.) The magnitude of the current can be obtain from or IL = mg (3.) I mg λg = = (3.3) L

4 Example 4: Charged particles in field Particle A with charge q and mass m A and particle with charge q and mass m, 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 are R and R, respectively. The direction of the magnetic field is perpendicular to the velocity of the particle. What is their mass ratio? The kinetic energy gained by the charges is equal to which yields mv qv = (4.) qv v = (4.) m The charges move in semicircle, since the magnetic force points radially inward and provides the source of the centripetal force: mv r = qv (4.3) The radius of the circle can be readily obtained as: which shows that r is proportional to mv m qv mv r = = = (4.4) q q m q / ( m/ q). The mass ratio can then be obtained from r ( m / q ) R ( m / q) = = (4.5) r m q R m q / / A A A A / / ( / ) ( / ) which gives ma m = 8 (4.6) 3

5 Example 5: ar 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? The magnetic force acting on a small differential current-carrying element Id s on the ring is given by df= I ds, where is 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 s is The total force acting on the ring then becomes df = df sin = ( Ids)sin (5.) y F = df = Isin ds= πrisin y y (5.) The force points in the +y direction and therefore is repulsive. 4

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