Problem 1:

As in the figure, an infinitely long wire carries a current I = 1.0 A.  It is bent so as to have a semi-circular detour around the origin, with radius 1.0 cm.  Calculate the magnetic field B at the origin.

Solution:

• Concepts:
  The Biot-Savart law
• Reasoning:
  The Biot-Savart law is used to find the field of an arbitrary current distribution.
• Details of the calculation:
  dB(r) = (μ₀/4π)Idl × (r - r')/|r - r'|³.
  For the field produced by the current flowing along the circular arc with radius b = 1 cm we have
  B(0) = (μ₀/4π)Iπb/b² k = (μ₀/4)I/b k, where k is the unit vector pointing into the page.
  The field produced by the straight-line segments at the origin is zero.
  The field at the origin therefore is B(0) = (μ₀/4)(1A)/(0.01 m) = 3.14*10^-5 T k.

Problem 2:

Consider a parallel-plate capacitor with circular plates of radius R and separation d << R.
The capacitor is being charged by a time-dependent current I(t) = I₀ sin(ωt).
(a)  Use Ampere's law to compute the displacement current density between the plates as a function of radius r (for r < R).
(b)  Find the magnetic field B(r, t) inside the capacitor (between the plates).
(c)  Discuss the behavior of B(r, t) for r > R, assuming the current is confined within the radius R.

Solution:

• Concepts:
  Maxwell's equations in quasi-static situations
• Reasoning:
  
• Details of the calculation:
  (a)  Chose the coordinate system so the capacitor plates lie in the xy-plane.
  We assume that the electric field between the plates is homogeneous.  E = E(t)k.
  Let I be positive if it flows in the z-direction, towards the right.
  The charge on the capacitor is Q = ∫ I(t) dt = -(I₀/ω) cos(ωt) + constant, where the constant depends on the initial condition.  Choose the constant to be zero.
  The capacitance is C = ε₀πR²/d = Q/V = Q/Ed.  E(t) = Q/( ε₀πR²) = -I₀cos(ωt) /(ωε₀πR²).
  Between the plates we have
  ∇×B = (1/c²)∂E/∂t.
  (1/c²)∂E/∂t = μ₀j_D, where j_D is the displacement current density.
  ∂E/∂t = I₀sin(ωt) /(ε₀πR²),  j_D = I₀sin(ωt) /(πR²) k
  
  (b)  ∮_Γ B∙dr =  (1/c²)∫_A∂E/∂t∙n dA .  Symmetry simplifies this equation.
  2πrB(r) = (πr²/c²) ∂E/∂t,  B = B(r) e_φ.
  ∂E/∂t = I₀sin(ωt) /(ε₀πR²).
  B(r) = rI₀sin(ωt)/(c²ε₀R²2π) = μ₀rI₀sin(ωt)/(2πR²).
  
  (c)  Outside the plates we have 2πrB(r) = (πR²/c²) ∂E/∂t,  B = B(r) e_φ.
  B(r) = μ₀I₀sin(ωt)/(2πr).
  For r > R the magnetic field is the same as the that produce by I(t) flowing in a ling straight wire.

Problem 3:

(a)  Find the magnetic field at a distance s from the long straight wire, carrying a steady current I.
(b)  Find the flux of B through a square loop of wire (side a) that lies on a table, a distance s from a very long straight wire with a current I, as shown in the figure.

(c)  If someone now pulls the loop directly away from the wire, at speed v, what emf is generated?  In what direction (clockwise or counterclockwise) does the current flow?

Solution:

• Concepts:
  Ampere's law, induced emf
• Reasoning:
  We can treat this as a Faraday's law problem or a motional emf problem.
• Details of the calculation:
  (a)  Let the current flow in the +z direction.
  Ampere's law (∮_ΓB∙dr = μ₀I_{through Γ}) yields B(r) = μ₀I/(2πr) e_φ.  (r is a cylindrical coordinate.)
  
  (b)  Flux Φ = ∫B∙ndA.  Let n point out of the page.
  Φ = [μ₀Ia/(2π)]∫₀^adx (s + x)^-1 = [μ₀Ia/(2π)]ln[(s + a)/s].
  
  (c)  We can treat this as a Faraday's law problem or a motional emf problem.
  Faraday's law:  -dΦ/dt = emf. 
  dΦ/dt = (dΦ/ds)v = [μ₀Iav/(2π)][1/(s + a) - 1/s)].
  emf = μ₀Ia²v/(2πs(s + a)).
  The current flows counterclockwise.
  
  Or:
  
  Motional emf: 
  Force on a mobile charge in a wire moving with velocity v in a magnetic field B:  F = qv×B.
  Force on an electron in a lower wire: q_evB = q_evμ₀Ia/(2πs) towards left.
  Force on an electrons in an upper wire: q_evB = q_evμ₀Ia/(2π(s + a) towards left.
  Work done per unit positive charge in moving counterclockwise around the loop:
  emf =  [μ₀Iav/(2π)][1/s - 1/(s + a)] = μ₀Ia²v/(2πs(s + a)).
  The current flows counterclockwise.

Problem 4:

A thin ring-shaped wire of mass M carries uniform electrical charge Q.  The wire rotates about its center axis with rotational velocity ω.
(a)  What is the ratio of its magnetic dipole moment to its angular momentum?
(b)  Classically, the electron is considered a spinning charged sphere.  According to quantum mechanics, the angular momentum of the spinning electron is h/4π, with h the Planck constant.  What torque acts on the electron at an angle θ between the magnetic moment and an outer magnetic field B in terms of the ratio derived in (a)?  Explain how (a) applies.
(c)  Derive the semi-classical precession angular velocity for an electron in a 3 Tesla magnetic field.

Solution:

• Concepts:
  Torque on magnetic dipoles, precession
• Reasoning:
  The magnetic moment of the rotating ring is proportional to its angular momentum.
  τ = μ × B = dL/dt.
  
• Details of the calculation:
  (a)  Let ω = ω k.  The wire has a uniform line charge density λ = Q/(2πR). 
  If it rotates with angular speed ω, the current is I = λv = QωR/(2πR) =  Qω/(2π). 
  The magnetic dipole moment of this rotating wire is  μ = IπR² k = QωR²/2 k.
  The angular momentum of this rotating wire is L = I_momentω k = MR²ω k.
  μ/L = Q/(2M).
  
  (b)  For this part assume B points in the +z-direction.
  The torque acting on the electron is τ = μ × B = (μ/L) L × B = -(μB /L) k × L.
  τ = dL/dt = Ω × L.
  Ω = -(μB /L) k is the precession frequency.
  
  Classically. the magnetic dipole moment and the angular momentum of a spinning sphere can be found by treating the sphere as a collection of rings and using the principle of superposition, assuming that the charge and the mass are both uniformly distributed.
  For each ring we have μ/L = Q_ring/(2M_ring), so for the whole sphere we have
  μ/L = Q_sphere/(2M_sphere).
  Ω = -(Q_sphereB/(2M_sphere)) k.
  
  For the classical electron we expect Ω = (q_eB/(2m_e)) k.
  [Using the measured magnetic moment Ω = q_eB/m_e k.]
  
  (c)  Putting in numbers:  Ω = (1.6*10^-19 C)*(3 T)/(9.1*10^-31 kg) = 2.64*10¹¹/s for the classical electron, using the measured magnetic moment.