More Problems

Problem 1:

Suppose that a very long coaxial line is divided into three regions
(i)  current I into the page for 0 < r < a,
(ii)  current 0 for radius a < r < b,
(iii)  current I out of the page for b < r < c.
Assume each conductor to have a uniform current density.  Find B for
(a)  r < a,
(b)  a < r < b,
(c)  b < r < c,
(d)  r > c.

Solution:

• Concepts:
  Amperes law
• Reasoning:
  Because of the cylindrical symmetry we can find the magnetic field from Ampere's law alone.
• Details of the calculation:
  (a)  r < a:  B(r) = μ₀I(r²/a²)/(2πr) = μ₀Ir/(2πa²), 
  B(r) = B(r)t,
  t is a unit vector tangent to any concentric circle, and points clockwise.
  (b)  a < r < b:  B(r) = μ₀I/(2πr),  B(r) = B(r)t.
  (c)  b < r < c:  B(r) = μ₀I/(2πr) - μ₀I(r² - b²)/(c² - b²))/(2πr)
  = (μ₀I/(2πr))((c² - r²)/(c² - b²)),    B(r) = B(r)t.
  (d)  r > c:  B(r) = 0.

Problem 2:

(a)  Compute the mutual inductance between two single turn circular loops of radii R and r, where R >> r.  The small loop is on the axis of the large loop at a distance Z from its center and in a plane parallel to the plane of the large loop.
(b)  How does the mutual inductance vary with the angle between the axes of the two loops?

Solution:

• Concepts:
  Flux F = ∫_A B∙dA ,  F = MI,  M = mutual inductance
• Reasoning:
  To find the mutual inductance, we calculate the flux through the small coil due to a current in the large coil.
  [When calculating the mutual inductance, you can calculate the flux through circuit 1 due to a current in circuit 2 or the flux through circuit 2 due to a current in circuit 1.  In a given problem, one of these calculations is often much simpler than the other.]
• Details of the calculation:
  (a)  The field on the axis of a current loop of radius R is B = k μ₀IR²/[2(R² + z²)^3/2] (SI units), if the current flows in the φ direction. 
  The flux through the loop of radius r is F = ∫_A B∙dA ,  F = πr² μ₀IR²/[2(R² + Z²)^3/2]. 
  (Since r << R, B is nearly constant over the area if the small loop.)
  F = MI,  M = πr²R²μ₀/[2(R² + Z²)^3/2].
  (b)  M = πr²R²μ₀/[2(R² + Z²)^3/2] cos(θ), where θ is the angle between the axes of the two loops.

Problem 3:

Calculate the magnetic dipole moment m of a disk of radius R and thickness d containing a uniform volume charge density ρ(r) = ρ₀r/R and rotating with angular velocity ω = ω k about its symmetry axis.  (Here r denotes the perpendicular distance from the rotation axis.)

Solution:

• Concepts:
  The magnetic dipole moment
• Reasoning:
  The magnetic dipole moment of the disk can be found by treating the disk as a collection of rings and using the principle of superposition. 
• Details of the calculation:
  Consider the ring between r and r + dr.
  dI = j(r)*dr*d,  j(r) = ρ(r)*v(r),  v(r) = r*ω,  dI = ρ(r)*ω*r*d*dr.
  dm = dIπr² k  = π*ρ(r)*ω*r³*d*dr =  π*ρ₀*ω*r⁴*d*dr/R
  m = ∫dm = k (1/R)π*ρ₀*ω*d*∫₀^R r⁵ dr = k π*ρ₀*ω*d*R⁴/5.

Problem 4:

A conducting rod of length L = 10 cm is placed on top of two conducting tracks.  The electrical potential difference between the tracks is U₀ = 15 V.  The resistance of the rod is R = 0.1 Ω.  The rod is tied to a mass of m = 1.2 kg with a thread that is redirected with a castor as shown in the figure.  The setup is inside a uniform magnetic field of B₀ = 1 T pointing upward.  The rod moves with constant speed.
(a)  Calculate the speed of the (weightless) rod.
(b)  What fraction of the electrical power delivered by the battery is converted into mechanical power?
(c)  At what resistance R does the rod remain stationary?

Solution:

• Concepts:
  Motional emf
• Reasoning:
  The conducting rod is moving in a plane perpendicular to B.
• Details of the calculation:
  Let the x-axis point to the right and let I be positive if it flows counterclockwise.
  (a)  Current in the circuit:  I = (V + ε)/R, ε = -BLdx/dt = -BLv, with v being positive if the rod moves towards the right.
  Force on the rod and mass: F = ILB - mg.  If F is positive, the rod is pulled towards the right.
  Constant speed:  F = 0,  mg = (V - BLv)BL/R,  v = V/(BL) - mgR/(BL)².
  v = -(1.2*9.8*0.1/(0.1)² + 15/0.1) m/s = 32.4 m/s.  The rod moves towards the right with speed 32.4 m/s.
  (b)  Mechanical power:  mgv = 381 W.  (The weight rises at a constant rate.)
  Electrical Power:  IV = V(V - BLv)/R = 1764 W.  (Work done by the battery per second.)
  Fraction of electrical power converted into mechanical power:
  (381 W)/(1764 W) = 21.6%.
  Power dissipated:  I²R = 1383 W.
  (c)  v = 0, R = VBL/mg = 0.128 Ω.

Problem 5:

Find the magnetic field B at the center of a flat spiral with current I.  The spiral is contained between two radii r and R and has N turns.  Do not consider the effect of the connecting wires.

Solution:

• Concepts:
  The Biot-Savart law
• Reasoning:
  For filamentary currents we have  B(r) = (μ₀/(4π))∫I dl' × (r-r')/|r-r'|³.   (SI units).
• Details of the calculation:
  Place the center of the spiral at the origin.  Assume the current flows counterclockwise.
  Let D = (R - r)/N be the spacing between turns and let the spiral starts at φ = 0.  (0 ≤ φ ≤ N2π)
  Consider a section of the spiral a distance r' = R - (D/2π)φ from the center of length dl = r'dφ. 
  The magnetic field at the origin due to this section is
  dB = (μ₀/(4π))I dφ /r' = (μ₀I/(4π))dφ/(R - (D/2π)φ). (Biot-Savart law)
  The magnetic field produced by the spiral at the origin is
  B = (μ₀I/(4π))∫₀^N2πdφ/(R - (D/2π)φ) = (μ₀I/(2D))ln(R)/ln(R - ND)
  = (μ₀IN/(2(R - r)))ln(R/r).
  B points perpendicular to the plane of the spiral.  If the current flows counterclockwise, B points out of the page.
  
  or, if N >> 1:
  For a single loop the field at the center of the loop of radius r' is B = μ₀I/(2r'), normal to the plane of the loop.
  Let I = kdr', k = NI/(R - r).  Here k is the magnitude of the surface current density.
  Then for the spiral we have
  B = μ₀∫_r^Rkdr'/(2r') = (μ₀k/2) ln(R/r) = (μ₀IN/(2(R - r)))ln(R/r).