Magnetic forces and the magnetic field, Lecture notes of Physics

Download Magnetic forces and the magnetic field and more Lecture notes Physics in PDF only on Docsity! Magnetic forces and the magnetic field December 19, 2015 Here we develop the Lorentz force law, and the Biot-Savart law for the magnetic field. 1 The Lorentz force law Like electrostatics, magnetostatics begins with the force on a charged particle. 1.1 The force law A beam of electrons passing a permanent magnet is deflected in a direction perpendicular to the velocity. Moving electrons – a current in a wire – deflects a compass needle brought nearby. These observations show that electric and magnetic phenomena influence one another. Careful measurement shows that in the presence of both electric and magnetic fields, the force on a charge Q is F = Q (E + v ×B) This is the Lorentz force law. Notice that this is consistent with the deflection described above since the cross product v ×B is always perpendicular to both v and B. Example: Motion of a particle in a constant magnetic field A particle of charge Q with initial velocity v0 moves in a constant magnetic field of magnitude B0. Find the motion. Choose the z-axis along the direction of the constant field, so that B = B0k̂, and choose the x-direction so that the initial velocity is v0 = v0x î + v0zk̂. The Lorentz law gives the force. Substituting the magnetic force into Newton’s second law, gives Qv ×B = m dv dt Writing out the separate components, Q (vxBy − vyBx) = m dvz dt Q (vyBz − vzBy) = m dvx dt Q (vzBx − vxBz) = m dvy dt Dropping Bx = By = 0, we have dvz dt = 0 QvyBz = m dvx dt −QvxBz = m dvy dt 1 The first equation immediately integrates twice to give z = z0 + v0zt so the charge moves with constant velocity in the direction parallel to the field. For the remaining two equations, differentiate each with respect to t, Q dvy dt Bz = m d2vx dt2 −Qdvx dt Bz = m d2vy dt2 and substitute the original equations to eliminate the first derivative terms and separate vx and vy, d2vx dt2 + Q2B2 z m2 vx = 0 d2vy dt2 + Q2B2 z m2 vy = 0 The solutions are clearly harmonic. Define the cyclotron frequency, ω ≡ QBz m so that vx = A sinωt+B cosωt vy = C sinωt+D cosωt These must satisfy the original equations dvx dt = ωvy dvy dt = −ωvx with the given initial conditions. Therefore, we must have ωA = ωD −ωB = ωC ωC = −ωB −ωD = −ωA so that vx = A sinωt+B cosωt vy = −B sinωt+A cosωt Fitting the initial values, we see that vx0 = B and 0 = v0y = A. Therefore, the velocity in the xy-plane is a circle given by vx = v0x cosωt vy = −v0x sinωt The full motion is a spiral parallel to the z-axis. 2 2.1 Conservation of charge Suppose we have a region of space with charge density ρ. Let some or all of this charge move as a current density, J. Now, since we find that total charge is conserved, we know that the total charge in some volume V can only change if the current carries charge across the boundary S if V. Therefore, with the charge in the volume V given by Qtot = ˆ V ρd3x the time rate of change of Qtot must be given by the total flux J across the boundary. Let n̂ be the outward normal of the boundary S of V. Then dQtot dt = − ˛ S J · n̂d2x (1) On the left side, we rewrite dQtot dt by interchanging the order of integration and differentiation , dQtot dt = d dt ˆ V ρd3x = ˆ V ∂ρ ∂t d3x while on the right we use the divergence theorem, − ¸ S J · n̂d 2x = − ´ V∇ · Jd 3x. Substituting both these changes into eq.(1), we have ˆ V ( ∂ρ ∂t + ∇ · J ) d3x = 0 Since the final equation holds for all volumes V it must hold at each point, leading us to the continuity equation: ∂ρ ∂t + ∇ · J = 0 (2) An equation of this sort holds anytime there is a conserved quantity. We define a steady state current to be one for which ρ and J are independent of explicit time dependence, ∂ρ ∂t = 0, ∂J∂t = 0. For a steady state current, the current density has vanishing divergence, ∇ · J = 0. 2.2 Examples of current density a) A wire of rectangular cross section with sides a and b carries a total current I. Find the current density if the flow is uniformly distributed through the wire. The current density is defined so I = dq dt = ¨ J · ŝd2x Taking the z-direction for the direction of the current, and noting that J is constant across the cross-section, we have Ik̂ = J k̂ aˆ 0 dx bˆ 0 dy I = Jab 5 and therefore J = I ab k̂ If we want J to be defined everywhere, we may use theta functions to limit the range, J = I ab k̂Θ (a− x) Θ (x) Θ (b− y) Θ (y) b) A thick disk of radius R and thickness L has a uniform charge density throughout its volume so that the total charge is Q. The disk rotates with angular velocity ω. What is the current density inside the disk? What is the current flowing between radius r and r + dr? What is the total current? First, the charge density in the disk is the total charge over the total volume of the disk, ρ = Q πR2L The velocity at radius r is ωrϕ̂, so J = ρv = Qrω πR2L ϕ̂ The current at radius r through a cross-section of width dr and height L is the integral dI = L̂ 0 Qrω πR2L drdzϕ̂ · ϕ̂ = Qωrdr πR2 The total current follows if we integrate across the radius of the cross-section, I = R̂ 0 dI = R̂ 0 Qωrdr πR2 = QωR2 2πR2 = Qω 2π This makes sense, since ω 2π is the fraction of a circle per unit time that the disk rotates, so we have that fraction of the total charge passing per unit time. 3 The Biot-Savart law 3.1 The magnetic field due to a given current density Next, we consider the effect of a current in producing a magnetic field. The experimental results are sum- marized for steady state line currents by the Biot-Savart law, B (x0) = µ0 4π ˆ J (x)× (x0 − x) |x0 − x|3 d3x 6 The constant µ0 is the permeability of free space with the value µ0 = 4π × 10−7N/A2. There is a clear parallel with our equation for the electric field. Instead of simply the charge density times the factor x0−x |x0−x|3 , we now require the cross product with the current density. We can simplify this expression in the case of a current carrying wire. Let the current density be restricted to a wire in the x-direction so that J (x) = îIδ (y) δ (z) Then B (x0) = µ0 4π ˆ îIδ (y) δ (z)× (x0 − x) |x0 − x|3 d3x = µ0 4π ˆ I (̂ idx ) × (x0 − x) |x0 − x|3 This is now integrated along the length of the wire. More generally, let the current be in the direction dl. Then we have the Biot-Savart law for a current, B (x0) = µ0 4π ˆ Idl× (x0 − x) |x0 − x|3 where the currents move along wires with tangents dl at positions x. This was the original form of the law. The law may also be specialized to surface current density K (x) by simply restricting the current density to a surface: B (x0) = µ0 4π ˆ K (x)× (x0 − x) |x0 − x|3 d2x Now that we have an expression for the magnetic field in terms of source current density, we can find the laws of magnetostatics. 3.2 Magnetic field of a long straight wire Find the magnetic field above a long straight wire carrying a steady current I. Using the Biot-Savart law for a steady current, we have B (x0) = µ0 4π ˆ Idl× (x0 − x) |x0 − x|3 Using cylindrical coordinates, let the current flow in the z-direction, so that dl = dzk̂. Then x = zk̂ and we take x0 = ρρ̂ so that the observation point is a distance ρ radially out from the wire. Evaluating the integrand, we have B (x0) = µ0 4π ∞̂ −∞ Idzk̂× ( ρρ̂− zk̂ ) ∣∣∣ρρ̂− zk̂∣∣∣3 = µ0I 4π ∞̂ −∞ dzρk̂× r̂∣∣∣ρρ̂− zk̂∣∣∣3 = µ0Iρ 4π ϕ̂ ∞̂ −∞ dz (ρ2 + z2) 3/2 Notice that the direction of the field circulates around the wire. Letting z = ρ tan θ and changing variables, dz = ρ cos2 θ dθ 7