PHYS-541 Chapter 13
Consider the two infinitely long straight currents shown in Figure 13-4. I' coincides with the y-axis. I is parallel to the yz-plane, is at a distance ρ from it, crosses the x-axis at y=z=0, and makes the angle α with the xy-plane as shown. Show that the force on I of C due to I' of C' is -1/2 µ₀II'cotαx-hat.
(13-16) F_C'→C = µ₀/4π ∫C∫C' [IdS x (I'dS' x R-hat)]/R² From the diagram: r = ρx-hat + yy-hat + zz-hat r' = y-hat y' R = ρx-hat + y-hat(y-y') + zz-hat dS = y-hat dy + z-hat dz dS' = y-hat dy' Find dS' x R: = -ρdy'z-hat + zdy'x-hat Find dS x (dS' x R): = -x-hatρdydy' + y-hatzdzdy' - z-hatzdydy' Plug into equation: F_x = -µ₀II'/4π ∫ ρdydy'/[ρ²+z²+(y-y')²]^3/2 Use tan substitution to solve y-hat and z-hat terms are 0.
How do we do Ampere's Law with several circuits acting on C?
F_C = F_total on C = ∑F_C'→C Use Ampere's Law for each circuit and add them up
How does F_C'→C compare to F_C→C'?
F_C'→C = -F_C→C
Apply (13-6) to the system of Figure 13-2 and thus show directly that (13-12) is obtained. (13-6) F_C'→C = -µ₀II'/4π ∫C∫C' dS•dS'/R² R-hat (13-12) F_C'→C = -µ₀II'ρ-hat/2πρ ∫ dz
From the diagram: r = ρρ-hat + zz-hat r' = z'z-hat R = r - r' = ρρ-hat + z-hat(z-z') R² = ρ² + (z-z')² dS = z-hat dz dS' = z-hat dz dS•dS' = dzdz' Plug into (13-6): F_C'→C = -µ₀II'/4π ∫C∫C' [ρρ-hat + z-hat(z-z') dzdz']/ [ ρ²+(z-z')² ]^3/2 split into two integrals: = -µ₀II'/4π ∫C∫C' [ρρ-hat dzdz']/ [ ρ²+(z-z')² ]^3/2 -µ₀II'/4π ∫C∫C' [z-hat(z-z') dzdz']/ [ ρ²+(z-z')² ]^3/2 By symmetry, the z component integral is 0: -µ₀II'/4π ∫C∫C' [ρρ-hat dzdz']/ [ ρ²+(z-z')² ]^3/2 Use u-substitution: u = z'-z, du=dz' →∫ du / (ρ²+u²)^3/2 = 2/ρ² →-µ₀II'/4π ∫ ρρ-hat dz [2/ρ²] = -µ₀II'/4π [2ρρ-hat/ρ²] ∫ dz = -2µ₀II'ρρ-hat/4πρ² ∫ dz = -µ₀II'ρ-hat/2πρ ∫ dz
Four very long straight wires each carry current of the same value I. They are all parallel to the z-axis and intersect the xy-plane at the points (0,0), (a,0), (a,a), and (0,a). The first and third have their current in the positive z-direction; the other two have currents in the negative z-direction. Find the total force per unit length on the current corresponding to the point (a,a).
We want to find the force per unit length on 4. F₄ = F₂→₄ + F₃→₄ + F₁→₄ = µ₀I²/2πa x-hat + µ₀I²/2πa y-hat + µ₀I²/2πa√2 *[x-hat cos π/4 + y-hat cos π/4] = µ₀I²/4πa * [2y-hat + 2x-hat - x-hat - y-hat] = µ₀I²/4πa * [x-hat + y-hat]
What does a steady current mean?
charges are moving with constant average speed
Ampere's Law for force between current elements
dF'_e'→e = -µ₀/4π [ (Ids) • (I'ds') ]R-hat / R²
What does Ampere's Law give us?
force between (steady) electric currents
A circular loop of radius a lies in the xy-plane with its center at the origin. It carries a current I' that circulates counterclockwise as seen when looking back at the origin from positive values of z. A very long current I is parallel to the x-axis, is going in the positive x-direction, and intersects the positive z-axis at a point a distance d from the origin. Find the total force on the circuit C carrying I.
r = x x-hat + d z-hat r' = a ρ-hat R = r-r' = x x-hat + d z-hat - a ρ-hat R² = x² + d² + a² -2axcosφ dS = dx x-hat dS' = a dφ' φ-hat Using the expression for F between two circuits: (13-6) F_C'→C = -µ₀II'/4π ∫C∫C' (dS•dS')R/R² dS•dS' = a dx dφ' x-hat•φ-hat = - a dx dφ' sinφ The y component of F_C'→C is given by: ρ-hat = x-hat cosφ + y-hat sinφ F_y = µ₀II'/4π ∫C∫C' -a²sin²φ'dφ'dx /(x²+d²+a²-2axcosφ')^3/2 = µ₀II'a²/4π ∫₀^2π sin² φ' * ∫inf dφ'dx / [(x-acosφ)²+d²+a²sin²φ')]^3/2 Use tan substitution to solve the second integral. Then, the x-hat and z-hat components are zero.
Example: two infinitely long parallel currents are carrying currents I and I' are a distance ρ. Use cylindrical coordinates and choose the z-axis to coincide with the source current I'. What is the R diagram? What is F_C'→C?
r = ρ ρ-hat + z z-hat r' = z' z-hat R = r-r' = ρ ρ-hat + (z-z') z-hat |R|² = ρ²+(z-z')² ds = dr = dz z-hat ds' = dr' = dz' z-hat [ds × (ds' × R-hat)]/R² = dz dz' z-hat × [z-hat × [ρρ-hat + (z-z') z-hat]]/R³ = - ρ dz dz' ρ-hat / [ρ²+(z-z')²]^3/2 F_C'→C = µ₀/4π ∫C∫C' (Ids × ( I'ds' × R-hat))/R² = -µ₀II'ρ ρ-hat/4π ∫dz ∫ dz'/[ρ²+(z-z')²]^3/2 after tan substitution: = -µ₀II' ρ-hat/2πρ ∫dz but this gives an infinite force, which isn't possible, so we introduce force per unit length: f_C = dF/dz = -µ₀II'ρ-hat/2πρ
