Problem 1:

A long, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is j.  The current density, although symmetrical about the cylinder axis, is not constant and varies according to the relationship
j = (b/r) e^(r-a)/δ k for r < a,  j = 0 for r > a,
where the radius of the cylinder is a = 5.00 cm, r is the radial distance from the cylinder axis, b is a constant equal to 600 A/m, and δ is a constant equal to 2.50 cm.
(a)  What is the total current I₀ passing through the entire cross section of the wire?
(b)  Using Ampere's law, derive an expression for the magnetic field B in the region r ≥ a. Express your answer in terms of I₀
(c)  Obtain an expression for the current I contained in a circular cross section of radius r ≤ a and centered at the cylinder axis.  Express your answer in terms of I₀, rather than b.
(d)  Using Ampere's law, derive an expression for the magnetic field B in the region r ≤ a.  Express your answer in terms of I₀, rather than b.
(e)   Evaluate the magnitude of the magnetic field at r = δ, a, and 2a.

Solution:

• Concepts:
  Ampere's law
• Reasoning:
  The problem has enough symmetry to find B using Ampere's law alone.
• Details of the calculation:
  (a)  I₀ = ∫j∙dA = ∫₀^a(b/r) e^(r-a)/δ 2πrdr = 2πb∫₀^a e^(r-a)/δ dr = 2πbδ(1 - e^-a/δ) = 81.5 A.
  (b)  Ampere's law:  ∮_ΓB∙dr = μ₀I_{through Γ}.
  For r > a we have 2πrB = μ₀I₀.  B = μ₀I₀/(2πr) e_φ.
  (c)  I(r) = 2πb∫₀^r e^(r-a)/δ dr = 2πbδ(e^(r-a)/δ - e^-a/δ)
  = 2πbδe^-a/δ(e^r/δ - 1) = I₀(e^r/δ - 1)/(e^a/δ - 1).
  (d)  2πrB = μ₀I₀(e^r/δ - 1)/(e^a/δ -1).  B = μ₀I₀(e^r/δ - 1)/[2πr(e^a/δ -1)] e_φ.
  (e)  B(δ) = μ₀I₀(e - 1)/[2πδ(e^a/δ -1)] = 1.75*10^-4 T.
  B(a) = μ₀I₀/(2πa) = 3.26*10^-4 T,  B(2a) = μ₀I₀/(4πa) = 1.63*10^-4 T.

Problem 2:

A copper rod slides on frictionless rails in the presence of a constant magnetic field B = B₀k.  At t = 0 the rod is moving in the y-direction with velocity v₀.
(a)  What is the subsequent velocity of the rod, if σ is its conductivity and ρ_m is the mass density of copper?  (Assume that the resistance of the rails can be neglected.)
(b)  For copper σ = 59.6*10⁶ (Ωm)^-1 and ρ_m = 8.9 g/cm³.  If B₀ is 1 gauss, estimate the time it takes the rod to stop.
(c)  Show that the rate of decrease of the kinetic energy of the rod per unit volume is equal to the ohmic heating rate per unit volume.

Solution:

• Concepts:
  Motional emf,  Newton's second law
• Reasoning:
  We have a well-defined filamentary circuit which changes its shape. 
  For such circuit we may write 
  ε = -d/dt∫_AB∙n dA.
• Details of the calculation:
  (a)  The rod and the rails form a circuit.  The flux through this circuit if F = B₀A.  A is the area of the circuit.
  Then ε = -dF/dt = -B₀w(dy/dt) = -B₀w v. 
  Here w is the length of the rod.
  I = |ε|/R,  R = w/(σa),  where a is the cross-sectional area of the rod.
  I = B₀σa v,  the current flows in the -φ direction.
  The force on the rod is F = -IwB₀ j = -B₀²σaw v j.
  F = ma = mdv/dt.
  ρ_maw dv/dt = -B₀²σaw v,  dv/dt = -B₀²σv/ρ_m,  v = v₀exp(-t/τ),
  τ =  ρ_m/B₀²σ.
  (b)  τ = (8.9*10³ kg/m³)/[(10^-4 T)²  59.6*10⁶ (Ωm)^-1] = 1.5*10⁴ s.
  When t = nτ, we may consider the rod to be at rest.  It is a judgment call what n should be.
  (c)  Kinetic energy T = ½Mv² = ½ρ_mV v₀²exp(-2t/τ).
  V = volume of the rod = aw.
  (1/V)dT/dt = -(ρ_mv₀²/τ)exp(-2t/τ).
  The rate of ohmic heating per unit volume is P/V = (1/V)ε²/R.
  P/V = (1/V)ε²σa/w = (1/V)B₀²w v²σa = (1/V)(ρ_m/(τσ))w v²σa
  = (1/V)(ρ_m/τ)w v²a = (ρ_m/τ)v² = (ρ_mv₀²/τ)exp(-2t/τ).
  (1/V)dT/dt = -P/V.

Problem 3:

A magnetic dipole p is located at the origin.  It lies in the x-z plane and makes an angle θ with the z axis.  At t = 0 a charge q is located at x = -x₀ and is moving with velocity v₀ in the positive y-direction.
Find the force the dipole exerts on q at t = 0.
 

Solution:

• Concepts:
  The dipole field, the Lorentz force, F = qv × B
• Reasoning:
  The magnetic field ue to the dipole is  B(r) = (μ₀/(4π))[3(m∙r)r/r⁵ - m/r³].
• Details of the calculation:
  Rotate your coordinate system as shown. 
  
  Then the dipole points in the z-direction and we can write  B = (mμ₀/(4πr³))[2cosθ' e_r + sinθ' e_θ')].
  At r = x₀ and θ' = θ + π/2 we have B = (mμ₀/(4πx₀³))[-2sinθ e_r + cosθ e_θ')].
  The force on the moving charge is F = qv × B.
  v = -v₀ e_φ,   F = (qv₀mμ₀/(4πx₀³)) [2sinθ e_θ' + cosθ e_r)].
  In terms of the original coordinates we have
  F = (qv₀mμ₀/(4πx₀³)) [-2sinθ k - cosθ i)].
  
  Aside:
  What if the problem asked for the force q exerts on the dipole?
  By Newton's 3^rd law F₁₂ = -F₂₁.
  The force q exerts on the dipole is F = (qv₀mμ₀/(4πx₀³)) [2sinθ k + cosθ i)].
  
  What if you take a different approach?
  The potential energy of an intrinsic magnetic dipole in a magnetic field is U = -m∙B.
  The force on the dipole on an external magnetic field is F = -∇U = ∇(m∙B) = (m∙∇)B, since m is constant.
  
  At t = 0 put q at the origin and the dipole on the x-axis at x = x₀.
  The magnetic field produced by the moving charge in the x-z plane (for x > 0) is B = (qv₀μ₀/(4πr²))e_θ.
  F = (m_x∂B/∂x + m_z∂B/∂z)|_{x=x0, y=0, z=0}.
  While on the positive x-axis B points in the -z-direction, we need to know B in the neighborhood of the x-axis to evaluate the derivatives.
  B = (qv₀μ₀/(4πr²))(-k x/r + i z/r) = (qv₀μ₀/(4πr³)) (-xk + zi) = A(r) (-xk + zi).
  ∂B/∂x = (∂A/∂x)(-xk + zi) - Ak.   ∂B/∂z = (∂A/∂z)(-xk + zi) + Ai.
  ∂A/∂x = (qv₀μ₀/(4π))∂(x² + y² + z²)^-3/2/∂x) = -(3qv₀μ₀/(4πr⁵))x.
  ∂A/∂z = (qv₀μ₀/(4π))∂(x² + y² + z²)^-3/2/∂x) = -(3qv₀μ₀/(4πr⁵))z.
  ∂B/∂x|_{x=x0, y=0, z=0} = (3qv₀μ₀/(4πx₀⁵))x₀² k - (qv₀μ₀/(4πx₀³)) k = (2qv₀μ₀/(4πx₀³)) k.
  ∂B/∂z|_{x=x0, y=0, z=0} = (qv₀μ₀/(4πx₀³)) i.
  F = (qv₀μ₀/(4πx₀³)) [2m_x k + m_z i)] = (qv₀mμ₀/(4πx₀³)) [2sinθ k + cosθ i)].
  It is easier to remember the equation for the dipole field and then use Newton's third law.

Problem 4:

A particle of charge q and mass m moves in a region containing a uniform electric field
E = Ei pointing in the x-direction and a uniform magnetic field B = Bk pointing in the z-direction.

(a)  Write down the equation of motion for the particle and find the general solutions for the Cartesian velocity components v_i(t) in terms of B, E, q, and m.
Hint:  Let ζ = v_x + iv_y, and solve for ζ(t).
(b)  Solve for the position of the particle as a function of time if the particle is released from rest at t = 0.

Solution:

• Concepts:
  The Lorentz force, F = q(E + v × B) = mdv/dt
• Reasoning:
  dv_z/dt = 0, v_z = constant.  We have to solve coupled differential equations for v_x and v_y using the hint, ζ = v_x + iv_y.
• Details of the calculation:
  (a)  dv_x/dt = qv_yB/m + (qE/m)
  dv_y/dt = -qv_xB/m
  Let ζ  = v_x + iv_y,
  dζ/dt + i(qB/m)ζ = (qE/m).
  To a particular solution of the first-order inhomogeneous differential equation,
  dζ/dt + i(qB/m)ζ = (qE/m), we add the solution of the homogeneous differential equation,
  dζ/dt + i(qB/m)ζ = 0, to find the most general solution.
  
  An inhomogeneous solution:
  dζ/dt = 0, ζ = ζ₀,  ζ₀ = -i(E/B).
  homogeneous solution:
  ζ = Aexp(-iωt), A = arbitrary complex constant = |A|exp(iφ),  ω = qB/m.
  general solution:
  ζ = |A|exp(-iωt + φ) - i(qE/m).
  v_x(t) = |A|cos(ωt + φ),
  v_y(t) = -|A|sin(ωt + φ) - (E/B).
  v_z(t) = 0.
  (b)  v_x(t) = 0, v_y(t) = 0 -- > φ = -π/2, |A| = E/B.
  v_x(t) = (E/B)sin(ωt),  v_y(t) = (E/B)cos(ωt) - (E/B).
  x(t) = ∫₀^t v_x(t')dt' = (mE/(qB²))(1 - cos(qBt/m)),  
  y(t) = ∫₀^t v_y(t')dt' = -Et/B +  (mE/(qB²))(sin(qBt/m).

Problem 5:

(a)  Find the magnetic field at any point in the x-y plane for y > 0 due to a wire of length l carrying a current I from x = 0 to x = l along the x-axis.
(b)  Use the result of (a) to find the self inductance of a square current loop of side l.  You may leave your result as a definite integral, but the limits must be specified and the integrand must be in terms of the dimensions of the loop, constants, and the variables being integrated.
(c)  A conducting square current loop of side l with sides parallel to the x- and y-axis has resistance R and self inductance L.  It moves at a speed v in the +x-direction.  The loop passes through a magnetic field given by B = B₀ k, -l/2< x < l/2, B = 0 elsewhere.  Find the current in the wire as a function of time assuming that I = 0 a long time before the loop reaches the magnetic field.

Solution:

• Concepts:
  The  Biot-Savart law, self inductance, induced emf
• Reasoning:
  We find the magnetic field due to the wire using the Biot-Savart law.  Given the magnetic field we use F = LI to find L.
• Details of the calculation:
  (a)  r = xi + yj.  B(r) = (μ₀/(4π))∫ I dl'×(r - r')/|r - r'|³.
  dl'= dx' i,  r - r' = (x - x')i + yj,  |r - r'|³ = ((x - x')² + y²)^3/2.
  dl'×(r - r') = y dx' k.
  B(r) = k (μ₀I/(4π))y∫₀^ldx' ((x - x')² + y²)^-3/2
  = k (μ₀I/(4π))y∫_x-l^xdx'' (x''² + y²)^-3/2 = k (μ₀I/(4πy))x''/(x''² + y²)^½|_x-l^x
  = k (μ₀I/(4πy))[x/(x² + y²)^½ - (x - l)/((x - l)² + y²)^½].
  
  (b)  F = LI,  F = ∫_AB(r)∙n dA.
  
  Flux due to side 1: F₁ = ∫_AB(r)∙k dA
  = (μ₀I/(4π))∫₀^l∫₀^ldx dy y^-1[x/(x² + y²)^½ - (x - l)/((x - l)² + y²)^½].
  L = (μ₀I/π)∫₀^l∫₀^ldx dy y^-1[x/(x² + y²)^½ - (x - l)/((x - l)² + y²)^½],
  since the flux due to all sides if 4 times the flux due to 1 side.
  
  (c)  ε = -dF/dt.  F = BA + LI.  (We let I be positive if it flows counterclockwise.)
  ε = -B dA/dt - L dI/dt.
  Let the right edge of the loop be at -l/2 at t = 0. 
  For t < 0 the current I = 0.
  
  Then for 0 < t < l/v we have
  ε = -Blv - L dI/dt = IR,  dI/dt + IR/L + Blv/L = 0.  I = -Blv/R + C exp(-Rt/L).
  I(0) = 0,  the constant C = Blv/R.  I(t) = -(Blv/R)(1 - exp(-Rt/L)).
  
  For l/v < t < 2l/v we have
  ε = Blv - L dI/dt = IR,  dI/dt + IR/L - Blv/L = 0.   I = Blv/R + C exp(-Rt/L).
  I(l/v) = -(Blv/R)[1 - exp(-Rl/(Lv))] = Blv/R + C exp(-Rl/(Lv)).
  C = -(2Blv/R)exp(Rl/(Lv)) + Blv/R.
  I(t') = Blv/R - (2Blv/R)exp(-Rt'/L) + (Blv/R)exp(-Rl/(Lv))exp(-Rt'/L)
  = Blv/R[1 - 2 exp(-Rt'/L) + exp(-Rl/(Lv))exp(-Rt'/L)]
  = Blv/R[1 - (2 - exp(-Rl/(Lv))) exp(-Rt'/L)],
  with t' = t - l/v.
  
  For t > 2l/v we have dI/dt + IR/L = 0.  I(t'') = I(2l/v) exp(-Rt''/L) with t'' = t - 2l/v.
  I(2l/v) = Blv/R[1 - (2 - exp(-Rl/(Lv))) exp(-Rl/(Lv)].