PHYS-541 Exam 3

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.

Why might flux change?

1. B can change in time 2. circuit may change in shape or size, changing the area enclosed 3. circuit may be moving by translation or rotation 4. combo of above

Explain why a change is necessary to the definition of Φ when time-dependent fields are allowed. E = -∇Φ - ∂A/∂t

1. In time-dependent fields, ∇∧E = 0. We cannot define E as E = -∇Φ. 2.If ∇•B=0, define B in terms of a new potential: B = ∇∧A 3. Use Faraday's Law: E∧∇ = -∂B/∂t = -∂(∇∧A)/∂t 4. Then: ∇∧ (E + ∂A/∂t) = 0 E + ∂A/∂t = 0 5. The vector field can always be written as the gradient of a scalar potential, so: E + ∂A/∂t = -∇Φ 6. Solving for E: E = -∇Φ - ∂A/∂t

What does the differential form of Faraday's Law mean?

1. a change in B gives an electric field 2. E is not conservative over time; it just take work and energy to push charges around a circuit

Derive the differential form of Faraday's Law, ∇∧E = -∂B/∂t, from ε_ind = -dΦ/dt.

1. ε_ind in a closed circuit is NOT 0: ε_ind = -dΦ/dt = ∫c E_nonconservative • ds 2. E is made of conservative and nonconservative parts: E_total = E_C + E_NC 3. →∫c E_tot•ds = ∫c E_NC•ds = -dΦ/dt = -d/dt [∫s B•da] 4. Equate the first and last term: ∫c E•ds = -∫s ∂B/∂t•da 5. Apply Stoke's Theorem to LHS: ∫s (∇∧E)•da = -∫s ∂B/∂t•da 6. Equate the integrands: ∇∧E = -∂B/∂t

How can you solve A from B, if B is constant?

A = 1/2 B × r

vector potential from a moving point charge

A = µ₀/4π qv/R

An infinitely long cylinder has a circular cross section of radius an and its axis along the z axis. A steady current I is distributed uniformly over the cross section and is in the positive z-direction. Use (16-23) to find A everywhere. If A outside the cylinder is written in the form (16-33), what is A on the z-axis? (16-23) Φ_B = ∫S B•da = ∫(∇×A)•da = ∫C A•ds (16-33) A = z-hat µ₀I/2π ln(ρ₀/ρ)

A = µ₀/4π ∫v J/R d𝜏 A is in the same direction as J due to symmetry: J = J z-hat, A = A(ρ) z-hat Application of Ampere's Law ∫ B•ds =µ₀I and symmetry gives B: B (ρ>a) = µ₀I/2πρ φ-hat B (ρ<a) = µ₀I/2πρ (πρ²/ππa²) φ-hat = µ₀Iρ/2πa² φ-hat Using ∫c A•ds = ∫s B•da, choose a loop C in the ρz plane with sides parallel to z and ρ, C₀ and C₁, as shown: ∫c A•ds = (A₁-A₂)L = -AL For ρ<a: ∫s B•da = -µ₀I/2πa² ∫ ρ dρdz φ-hat • φ-hat = Lµ₀I/2πa² (ρ₂²-ρ₁²/2) = Lµ₀I/2πa² (ρ²/2) Equate the two: ∫c A•ds = ∫s B•da -A₁L = Lµ₀I/2πa² (ρ²/2) →A₁ = -µ₀I/4πa² ρ² + C For ρ>a: ∫s B•da = -µ₀I/2π ∫∫ dρ/ρ dz →(A₁₀-A₂₀)L = -µ₀IL/2π * [ln ρ]^ρ₁₀_ρ₂₀ = -µ₀IL/2π * [ln ρ₁₀ - lnρ₂₀] →A₀ = -µ₀I/2π ln(ρ) + C Using (16-33), assuming A=0 at ρ=ρ₀: A₀ = µ₀I/2π ln(ρ₀/ρ) A is continuous, so equate the two, solve for A₁, plug in ρ=0.

In Figure 15-14, we show a portion of a current-free region in which B is uniform. Show that such an induction cannot drop abruptly to zero, as is indicated schematically by the location of the arrows, by applying (15-1) to the dashed rectangular path shown. Show qualitatively that a change in B of the general nature shown in Figure 6-10 will be compatible with (15-1).

A. ∫B•dS = µ₀I Since there is no current, I=0, and: ∫B₁→₂ • dS + ∫B₂→₃ • dS + ∫B₃→₄ • dS +. ∫B₄→₁ • dS ∫B₁→₂ • dS: B and dS are at 180°, result is negative ∫B₂→₃ • dS: B and dS are at 90°, result is 0 ∫B₃→₄ • dS: B=0, result is 0 ∫B₄→₁ • dS: B and dS are at 90°, result is 0 The first term cannot be zero unless B is zero, so this situation isn't possible. B. If we allow fringing, terms 2, 3, and 4 will be positive, which allows for the possibility that ∫B•dS is zero.

Consider the infinitely long coaxial cylindrical conductors shown in Figure 6-12. The inner conductor carries a total current I in the z-hat direction, while the outer conductor carries a current I in the -z-hat direction. Assume the currents to be uniformly distributed over their respective cross-sections. Find B everywhere.

Ampere's Law is: ∫B•dS = µ₀I For each region, we use Ampere's Law around C, where C is a circular loop in the xy-plane, centered at the z-axis, of varying radius: dS = ϕ-hat ρdϕ By symmetry, any B field must have only a ϕ component (can also find by right-hand rule, putting your thumb in the direction of the current) B = B_ϕ ϕ-hat B_ϕ = B_ϕ(ρ) 1. r>c: ∫B•dS = 0, no current B = 0 2. b<r<c: ∫B•dS = µ₀[I - I (πρ²-πb₂/πc²-πb²)] 2πρB_ϕ = µ₀I/(c²-b²) (c²-ρ²) B = µ₀I/2πρ (c²-b²/c²-ρ²) ϕ-hat 3. a<r<b: ∫B•dS = µ₀I 2πρB_ϕ = µ₀I B = µ₀I/2πρ ϕ-hat 4. r<a: ∫B•dS = µ₀I(πρ²/πa²) 2πρB_ϕ = µ₀I(ρ²/a²) B = µ₀I/2π ρ/a² ϕ-hat

A long cylindrical nonmagnetic conductor of radius b has a coaxial cylindrical hole of radius a drilled along it, that is, it is like Figure 18-1 with conductor in region 2 and everything else a vacuum. It carries a current I distributed uniformly over the cross-section. Find the magnetic energy associated with the induction in a length L of the conductor.

Apply Ampere's Law to circles of radius ρ. from symmetry: B = B_ϕ(ρ) ϕ-hat 1. ρ < a: I_enc = 0 → B=0 2. a < ρ < b: I_enc = I (πρ²-πa² / πb²-πa²) = I (ρ²-a² / b²-a²) →∫B•ds = B_ϕ 2πρ = µ₀ I(ρ²-a²)/(b²-a²) B_ϕ = µ₀ I(ρ²-a²) / 2πρ(b²-a²) 3. ρ > b: ∫B•ds = B_ϕ 2πρ = µ₀ I B_ϕ = µ₀ I / 2πρ Magnetic energy is the sum of U_B in all 3 regions: U_B = ½ µ₀∫B² d𝜏 in cylindrical: d𝜏 = ρdρdϕdz 1. ρ < a: B = 0 → U_M = 0 2. a < ρ < b: B = µ₀ I(ρ²-a²) / 2πρ(b²-a²) U_M = 1/2µ₀ ∫ µ₀²I²/4π²(b²-a²) (ρ²-a²)/ρ ρdρdϕdz = µ₀I²L2π/8π²(b²-a²)² ∫_a^b (ρ²-a²)²/ρ dρ = µ₀I²L/4π(b²-a²)² [ (ρ⁴/4 2a²ρ²/2 + a⁴ln(b/a) ]_a^b = I²Lµ₀/4π(b²-a²)² [ (b⁴-a⁴)/4 + a⁴ - a²b² + a⁴ln(b/a) ] = I²Lµ₀/16π(b²-a²)² [ 4a⁴ln(b/a) + (b²-a²)(b²-3a²)] 3. ρ > b: not within cylinder

A toroidal coil of N turns has a central radius b and a square cross-section of side a. Find its self-inductance.

Assuming B_ρ and B_z are negligible: self-inductance L = NΦ_B / I, Φ_B is flux per turn Φ_B = ∫B•da To find B: apply Ampere's Law around C₁, inside the toroid ∫B•da = µ₀I_enc B_φ(2πρ) = µ₀NI B_φ = µ₀NI/2πρ φ-hat where ρ is between b-½a and b+½a Now finding Φ_B: Φ_B = ∫B•da, da=dρdz φ-hat = µ₀NI/2π ∫ dρ/ρ ∫ dz the bounds of the first integral are b-a/2 → b+a/2 the bounds of the second integral are 0→a = µ₀NI/2π a ln (2b+a)/(2b-a) Now finding L: L = NΦ_B / I = N/I * (µ₀NI/2π a ln (2b+a)/(2b-a) ) = µ₀N²Ia/2πI ln(2b+a)/(2b-a) = µ₀N²a/2π ln [(2b+a)/(2b-a)]

What is B with moving point charges?

B = µ₀/4π (q'v' × R-hat)/R²

What is B of an ideal infinite solenoid?

B = µ₀nI'

Example: infinite uniform plane current sheet We have an edge-on infinite plane sheet with K out of the page. Assume |K| = K = constant. What is B?

B is perpendicular to K, parallel to the plane of the sheet, and oppositely directed above/below the sheet. Choose a rectangle above/below the sheet as the path of integration. On the horizontal sides, B is perpendicular to ds, so B•dS = 0. The only contribution is from the vertical sides: B•ds = B(D)ds Since |K| is current per unit length, I_enc = KL, and: ∫C B • dS = µ₀ I_enclosed →2BL = µ₀KL →B = 1/2 µ₀K So B is independent of distance from the current sheet

Example: infinitely long straight current Assume the current I to be uniformly distributed over the circular cross section of radius a of an infinitely long cylinder. What is B?

By symmetry, B is in the plane perpendicular to I, and depends only on ρ. So: B = B_ϕ(ρ) ϕ-hat If we choose the path of integration as a circle of radius ρ: ds = ρ dϕ ϕ-hat ∫C B • dS = ∫₀^2π B_ϕ ϕ-hat • ρ dϕ ϕ-hat → 2πρB_ϕ = µ₀I_enc → B_ϕ = µ₀I_enc/2πρ 2 regions: outside cylinder and inside cylinder 1. outside cylinder: ρ>a, C encloses total current I I_enc = I B_ϕ(ρ) = µ₀I_enc/2πρ (ρ>a) 2. inside cylinder: ρ<a, C does not enclose total current; only fraction equal to circular area enclosed by C divided by cross section I_enc/I = πρ²/πa² I_enc = I(ρ²/a²) B_ϕ(ρ) = µ₀Iρ/2πa² (ρ<a)

Two infinitely long straight currents are parallel to the z-axis. One of them, carrying a current I₁, intersects the xy-plane at the point (x₁,y₁). The other, with current I₂, intersects the xy-plane at (x₂,y₂). Find the resultant B produced by them at any field point (x,y,z).

Due to I₁ at P, Biot-Savart Law: B₁ = µ₀I/4π ∫c dS'xR/R³ r = x x-hat + y y-hat + z z-hat r' = x₁x-hat + y₁y-hat + z'z-hat R = r-r' = x-hat(x-x₁) + y-hat(y-y₁) + z-hat(z-z') dS' = dz'z-hat dS'xR = dz'[y-hat(x-x₁) - x-hat(y-y₁)] Biot-Savart Law: B₁ = µ₀I/4π ∫c dS'xR/R³ = µ₀I₁/4π ∫ dz'[y-hat(x-x₁) - x-hat(y-y₁)] /[(x-x₁)² + (y-y₁)² + (z-z')²]^3/2 Use tan substitution to solve; solve B₂ in the same way (change x₁→x₂, y₁→y₂) B_TOT = B₁+B₂

In the circuit shown in Figure 14-10, the curved lines are semicircles with common center C. The straight portions are horizontal. At this instant, a point charge q located at C has a velocity v directed vertically downward. Find the magnetic force on q.

F_B,on,q = q(vxB) B = µ₀I/4π ∫c dS'xR/R³ Thus, we need to evaluate B at point C. This is a combination of the B due to two straight elements and two curved elements. Straight elements: dS' = x-hat dS' r = 0, q is at the origin r' = ±x' x-hat R = r-r' = ∓x' x-hat But dS' x R = 0, because both are along x-hat →B=0 For curved elements: dS' = ϕ-hat ρdϕ = ϕ-hat bdϕ, or -ϕ-hat adϕ' (outer/inner) r = 0 r' = ρ-hat b or ρ-hat a R = r-r' = ρ-hat b or -ρ-hat a dS' x R = ϕ-hat x ρ-hat[-b²]dϕ' or ϕ-hat x ρ-hat[a²]dϕ' = z-hat b²dϕ' or -z-hat a²dϕ' Biot-Savart Law: B = µ₀I/4π ∫c dS'xR/R³ →B = µ₀I'/4b z-hat or -µ₀I'/4a z-hat (integral is over π, not 2π) Thus, at C, due to both curved elements: B = µ₀I'/4 z-hat * (1/b - 1/a) = µ₀(a-b)I'/4ab z-hat Hence, F_B = q(v x B) where v=-v y-hat = -qv µ₀/4ab (a-b)I'(y-hat x z-hat) = -x-hat µ₀qvI'(a-b)/4ab But b>a, hence the direction is x-hat, and: F_B = µ₀qvI'(b-a)/4ab

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

A certain induction has the form B = (αx/y²)x-hat + (βy/x²)y-hat + f(x,y,z)z-hat, where α and β are constants. Find the most general possible form for the function f(x,y,z). Find the current density J and verify that it corresponds to a steady current distribution.

Finding f: ∇•B = 0 → ∂/∂x (αx/y²) + ∂/∂y (βy/x²) + ∂/∂z (f) = 0 → (α/y²) + (β/x²) + ∂f/∂z = 0 → ∂f/∂z = (-α/y²) - (β/x²) →f = -αz/y² - βz/x² + g Finding J: ∇ × B = µ₀J ∇ × B = =[x-hat, y-hat, z-hat; ∂/∂x, ∂/∂y, ∂/∂z; αx/y², βy/x², f] = x-hat (∂f/∂y - 0) - y-hat (∂f/∂x - 0) + z-hat (-2βy/x³ + 2αx/y³) = µ₀J Verifying steady current: ∇•J = 0 ∇•J = ∂/∂x (2αz/y³ + ∂g/∂y) - ∂/∂y (2βz/x³ + ∂g/∂x) + ∂/∂z (2αx/y³ - 2βy/x³) = ∂²g/∂x∂y - ∂²g/∂x∂y = 0

Suppose the current in the infinitely long straight current C' of Figure 13-5 is given by I = I₀e^-λt where I₀ and λ are constants. Find the induced emf that will be produced in the rectangular circuit. What is the direction of the induced current?

Finding induced emf: First use Ampere's Law to find B due to I' through C: ∫B•ds = µ₀I B(2πρ) = µ₀I₀e^-λt B = µ₀I₀e^-λt/2πρ Then finding flux through C: Φ = ∫s B•da, da=dρdz φ-hat = µ₀/2π I₀e^-λt ∫ ∫ dρdz/ρ first integral: 0→b second integral: d→d+a = µ₀/2π I₀be^-λt ln(d+a /d) Then induced emf: ε_ind = -dΦ/dt |ε_ind | = µ₀/2π b ln(d+a /d) I₀λe^-λt Direction of induced current: From Lenz's Law, B_induced and I_induced are the opposite direction of ∆B. →Flux through C is decreasing with time, so ∆B is -φ-hat →therefore, B_ind is φ-hat, and I_ind is clockwise

A uniform induction B is parallel to the axis of a nonmagnetic cylinder of radius an and dielectric constant κ_e. If the cylinder is rotated about its axis with a constant angular velocity ω parallel to B, find the polarization produced within the cylinder and the surface charge on a length L of it.

First finding polarization: Use (17-29): E' = E + v×B If E=0: E' = v×B v = ρω φ-hat, B = B z-hat →E' = ρωB(φ-hat × z-hat) = ρωB ρ-hat P = (κ_e - 1) ε₀ E = (κ_e-1)ε₀Bωρ φ-hat Now finding surface charge: surface charge density = σ = P•n-hat n-hat = ρ-hat P is on the surface (ρ=a) surface charge = Q = ∫σ•da = σl2πa = (κ_e-1)ε₀Bω2πa²l

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

Under what conditions, if any, will the contribution of region 1 of Figure 18-1 to the self-inductance of the coaxial line be greater than that of region 2? From (18-34): Contribution to L from region 1: µ₀L/8π Contribution to L from region 2: µ₀L/2π ln(b/a)

If 1>2: µ₀L/8π > µ₀L/2π ln(b/a) 1/8 > 1/2 ln(b/a) 0.25 > ln(b/a) e^0.25 > b/a b < ae^0.25 b < 1,28a

An infinite plane current sheet coincides with the xy-plane. Its surface density is K'=K' y-hat, where K'=constant. Another infinite plane current sheet is parallel to the xy-plane and intersects the positive z-axis at z=0. The second current density is K'=-K' y-hat. Find B everywhere.

If the field point is located at (x,y,z): For the sheet at z=0: r = x-hat x + y-hat y + z-hat z r' = x-hat x' + y-hat y' R = r-r' = x-hat(x-x') + y-hat(y-y') + z-hat(z) K' = k' y-hat Biot-Savart Law: B₁ = µ₀/4π ∫∫ (K' x R)/R³ da' K' x R = -z-hat K'(x-x') + x-hat K'z B₁ = µ₀/4π ∫∫ (K' x R)/R³ da' = µ₀/4π ∫∫ K'[x-hat z - z-hat(x-x')]dx'dy' / [(x-x')+(y-y')²+(z)²]^3/2 split into separate integrals the z-hat component integral is odd, so B_z=0 For x component: B₁ = x-hat µ₀K'/4π ∫ 2z/[(x-x')+z²] dx' use tan sub to solve The current sheet at z=d is same, just negative. Use tan sub to solve. So, between the sheets: B = B₁+B₂ Outside the sheets (z<0 or z>d): B = 0

Consider a very long cylindrical beam of charged particles. The beam has a circular cross section of radius a, a uniform charge density ρ_ch, and the particles have the same constant velocity v. Find B inside and outside of the beam and express your results in terms of these given quantities.

In general: J = ρ_ch v I = ∫ J • da = ρ_ch vπa² where a is the radius of a cylinder ∫B•dS = µ₀I The moving charges constitute a current I. Using the usual symmetry arguments and integrating around a concentric circle of radius ρ: 1. r>a: B_ϕ2πρ = µ₀ρ_ch vπa² B_ϕ = 1/2 µ₀a²ρ_ch v/ρ ϕ-hat But ϕ-hat = z-hat x ρ-hat = v x ρ/|v||ρ| B_ϕ = 1/2 µ₀a²ρ_ch v/ρ (v x ρ/|v||ρ|) = 1/2 µ₀ρ_ch(a/ρ)² (v x ρ) 2. r<a: I = ρ_ch vπa² πρ²/πa² = ρ_ch vπρ² B = 1/2 µ₀ ρ_ch ρ² v/ρ (v x ρ/vρ) = 1/2 µ₀ ρ_ch (v x ρ)

A sphere of radius a contains a total charge Q distributed uniformly throughout its volume. It is set into rotation about a diameter with constant angular speed ω. Assume that the charge distribution is not altered by the rotation and find A at any point on the axis of rotation.

Q = 4/3 πa³ρ_ch Rotating charges constitute a current, which creates a B-field, hence we have an A. For a point P on the z-axis: r = z z-hat r' = r' r-hat J = φ-hat ρ_ch |v| = φ-hat ρ_ch ωr'sinθ A = µ₀/4π ∫V J/R d𝜏 = µ₀/4π ∫ φ-hat ρ_ch ωr'sinθ r'² dr'sinθ'dθ' dφ' / (z² + r'² - 2zr'cosθ')^1/2 = ωµ₀ρ_ch/4π ∫∫∫ (dφ' r'²dr' (-x-hat sinφ' + y-hat cosφ)) sin²θ'dθ' / (z² + r'² - 2zr'cosθ')^1/2 the x and y-hat component integrals have the integrals ∫0→2π sinφ'dφ' and ∫0→2π cosφ'dφ', which are both 0, so: A = 0

A circular dielectric disk of radius a has a uniform surface charge density σ on it. It is rotated with constant angular speed ω about an axis that is normal to the surface of the disk and passes through its center. Assume that the charge is not affected by the rotation and find B at an arbitrary point on the axis of rotation. What is B at the center of the disk?

We have a rotating surface charge, which constitutes a surface current r = z z-hat r' = ρ ρ-hat K' = ϕ-hat σωρ' R = r-r' = z z-hat - ρ' ρ-hat = z z-hat - ρ'cosϕ x-hat - ρ'sinϕ y-hat K' = σv = σωρ' ϕ-hat = σωρ' [-x-hat sinϕ + y-hat cosϕ] B₁ = µ₀/4π ∫∫ (K' x R)/R³ da' da' = ρ'dρ'dϕ' K' x R = σωρ'[x-hat zcosϕ + y-hat zsinϕ + z-hat ρ'] →B = µ₀σω/4π ∫₀2π ∫₀a [x-hat zcosϕ + y-hat zsinϕ + z-hat ρ']ρ'²dρ'dϕ' / [z²+ρ'²]^3/2 The x and y components are zero because the integral is 0→2π and we have both cos and sin. B = z-hat = µ₀σω/4π 2π ∫₀a ρ'³dρ' / [z²+ρ'²]^3/2 Solve the integral using tan substitution At the center of the disk, z=0: →B = z-hat 1/2 µ₀σωa

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

A self-inductance L, a resistance R, and a battery of emf ε_b are all connected in series. Use energy considerations to show that the current I satisfies the differential equation L di/dt + Ri = ε_b. Now assume that I ≠ 0 and the battery is switched out of the circuit. Solve the resulting equation and find the relaxation time of this system.

energy provided by battery = energy stored in inductance + energy dissipated by resistor dU_batt = dU_M + dU_R ε_b*dq = ½L2IdI + I²Rdt ε_b*dq = LIdI + I²Rdt Divide by dt: ε_b*dq/dt = LIdI/dt + I²Rdt/dt ε_b*I = LIdI/dt + I²R ε_b = L dI/dt + IR If the battery is taken out of the circuit, at any time after: dU_M + dU_r = 0 →LIdI + I²Rat = 0 dI/dt = -IR/L ∫dI/I = - ∫ R/L dt ln I/I₀ = -Rt/L I/I₀ = e^(-Rt/L) I = I₀e^(-Rt/L) = I₀e^(-t/𝜏) 𝜏 = relaxation time = L/R

What vacuum valve of the inductance will result in a magnetic pressure of one atmosphere?

f_m = magnetic pressure = B²/2µ₀ →B² = f_m 2µ₀ 1 atmosphere = 1.01*10⁵ N/m² B = (1.01*10⁵ N/m²)(2)(4π*10-⁷)^1/2 = 0.504 Tesla

What does Ampere's Law give us?

force between (steady) electric currents

back emf

induced back emf is a resultant of voltage/current changing. the quicker the voltage changes, the larger back emf is induced. ε_ind = -dΦ/dt = -L dI/dt

What is B?

magnetic induction OR magnetic flux density OR B-field

Example: infinitely long ideal solenoid We have a current sheet of surface density K circulating around the cylinder. What is B?

n turns per unit length with current I means we have current per unit length nI, and: K = nI If the solenoid is infinitely long, B is parallel to axis B₀ = B outside solenoid B₁ = B inside solenoid Choose the rectangular path of integration. Only the vertical sides contribute, so: I_enc = KL = nIL ∫C B • dS = µ₀ I_enclosed →B₁L-B₀L = µ₀KL = µ₀nIL →B₁-B₀ = µ₀K = µ₀nI Which is independent of D. So, if we choose D→∞, B₀=0: B₁ = µ₀K = µ₀nI B₀ = 0 →B₁ = µ₀K z-hat = µ₀nI z-hat So, induction is confined entirely to the interior and is independent of radius.

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.

A square of edge 2a lies in the xy-plane with the origin at the center. The sides of the square are parallel to the axes, and a current I goes around it in a counterclockwise sense as seen from a positive value of Z. Find A at all points within the square. What is A at the center?

r = x x-hat + y y-hat r' = a x-hat + y' y-hat R = r - r' = x-hat (x-a) + y-hat (y-y') ds' = dy' y-hat A(r) = µ₀/4π ∫c' I'ds'/R = µ₀/4π ∫-a→a I'dy' y-hat / [(x-a)² + (y-y')²]^1/2 = -y-hat µ₀I'/4π dy'/ [(x-a)² + (y-y')²]^1/2 Use tan sub to integrate

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

What does the negative sign in Faraday's Law represent?

the direction of the induced emf compared to the direction of C

Lenz's Law

the induced emf, and by extension the induced current, opposes the change that is producing it B_induced is in the opposite direction as ∆B

A circle of radius a lies in the xy-plane with its center at the origin. Find the solid angle Ω subtended by this circle at a point on the positive z-axis.

Ω = ∫ da•R/R³ r = z z-hat r' = ρ' ρ-hat da = z-hat ρ'dρ'dϕ R = r -r' = z-hat z - ρ-hat ρ' |R| = (z²+ρ²)^1/2 dA • R = z-hat ρ'dρ'dϕ • z-hat z - ρ-hat ρ' = zρ'dρ'dϕ' Ω = ∫₀2π ∫₀a zρ'dρ'dϕ'/(z²+ρ²)^3/2 = 2πz ∫₀a ρ'dρ'/(z²+ρ²)^3/2 where I = ρ'dρ'/(z²+ρ²)^3/2 Solve via tan substitution, where: ρ'=ztanα, dρ'=zsec²αdα

Find the emf induced in the loop of Figure 17-4 when it is rotating so that ϕ=ωt+ϕ₀ while simultaneously B is oscillating at the same frequency; that is, B = B₀cos(ωt+α) x-hat. Is it possible to choose the constants ϕ₀ and α in such a way that induced emf is always zero?

ε_ind = -dΦ/dt Finding Φ: Φ = ∫(B•n-hat)da da=dzdy →Φ = abB₀cos(ωt+α) cos(ωt+φ₀) ε_ind = -dΦ/dt = abB₀[ωsin(ωt+α)cos(ωt+φ₀) + ωcos(ωt+α)sin(ωt+φ₀) ] = ωabB₀[sin(ωt+α)cos(ωt+φ₀) + cos(ωt+α)sin(ωt+φ₀) ] = ωB₀ab sin(ωt+α+ωt+φ₀) = ωB₀ab sin(2ωt+α+φ₀) The sine function can never be zero, for all t, for any value of α and φ₀

Apply (15-1) to a circle of radius ρ<a whose plane is perpendicular to the z-axis of the solenoid of Figure 15-11 and thus show that B_ϕ=0 in the interior. (15-1) ∫B•dS = µ₀I_enc

∫B•dS = µ₀I_enc dS = ϕ-hat ρdϕ For the loop C shown: ∫B • ϕ-hat ρdϕ = µ₀I The right-hand side is zero for this C because no current is passing "through" C →∫B_ϕ ρdϕ = 0 We know that B_ϕ is independent of ϕ by symmetry, so B_ϕ can come outside the integral, and ρ is constant throughout a circle: B_ϕ ρ ∫dϕ = 0 B_ϕ ρ 2π = 0 B_ϕ = 0 for all ρ

What is the flux of B through a closed surface?

∫S B • da = 0 The flux of B through any closed surface is always zero.