more derivations Flashcards
derive the equation for the force per unit length of 2 parallel current-carrying wires
the first wire will produce a magnetic field with B₁ =µ₀I₁/2πs (where s is the wire separation) wire 2 feels a force F₂ = B₁I₂L = µ₀I₁I₂L/2πs
the first wire experiences the force F₁ = F₂
the force per unit length is thus
F = µ₀I₁I₂/2πs
derive the B-field from an infinite wire using ampere’s law
draw a wire. put a loop of radius s around it
B =µ₀I/2πs (theta hat) for an infinite wire
dl = dl (theta hat)
hence B⋅dl = B*dl
since the radius, s, is constant B is also constant
B∮dl
=B * 2πs = µ₀I (which obeys Ampere’s law)
use a Ampere’s law on a solenoid to derive its B-field
sides a,b,c,d. side a–>b is inside the solenoid
∮B⋅dl = ∮B⋅dl (a–>b) + ∮B⋅dl (b–>c) + ∮B⋅dl (c–>d) + ∮B⋅dl (d–>a)
for b–>c and d–>a dl and B are perpendicular so B⋅dl =0
for c–>d B= 0 as it’s outside the solenoid
since B is uniform (constant) for a–>b, ∮B⋅dl = BL =µ₀* I (enclosed)
I (enclosed) = nLI
hence BL =µ₀nLI
so B = µ₀nI
where I is the current flowing through the wire that makes the solenoid
derive the e.m.f created by a homopolar generator
draw a conducting disk perpendicular to a magnetic field
for a small segment (from r –> r+dr)
the velocity is v= ωr (*theta hat)
the force per unit charge is v x B = vB r̂ (since v and B are perpendicular)
this = ωrB r̂
dε = (v x B) ⋅dl = ωrB dr
∴ ε = ½ωBR² (R: disks total radius)
derive the energy stored inside an inductor
dW = -ε*dq
I = dq/dt hence dq = I*dt
ε = -L dI/dt
dW = L dI/dt * I*dt
= LI dI
W = L * ∫I dI (limits from 0 to I)
= ½LI²
derive the mutual inductance for 2 solenoids
Two solenoids that share the same longitudinal axis. Both of length ‘l’ and cross-section ‘A’.
B₁ =µ₀n₁𝐼₁ = µ₀𝐼₁N₁/l (From Ampere’s law)
Φ₂ = M*𝐼₁
M = Φ₂/𝐼₁ = N₂B₁A / 𝐼₁
= µ₀𝐼₁N₁N₂A/l𝐼₁
= µ₀N₁N₂A/l
l: length
𝐼: current
derive the transformer equations
for two coils wrapped around an iron core:
Φ₂ ∝ BN₂ , Φ₁ ∝ BN₁
hence: Φ₁/Φ₂ = N₁/N₂
Φ₁ = N₁/N₂ * Φ₂
ε = dΦ/dt
hence: ε₁ = N₁/N₂ * ε₂
ε₁/ε₂ = N₁/N₂
power is conserved: (P=IV)
ε₁I₁ = I₂ε₂
derive the self-inductance of a solenoid
calculate the B-field inside using amperes law
∫B⋅dl = µ₀I(enc) (where I(enc) = Nl)
B = µ₀n*I (where n = N/l)
Then calculate the flux
Φ = N∫B⋅dA = NBA (B is constant)
for the inductance
Then use L = Φ/I = µ₀N²A/l