even more derivations Flashcards
derive the magnetic field energy density
for a solenoid:
B = µ₀nI
hence I = B/µ₀n (I: the current)
L = µ₀n²lA (l: the length)
and U = ½LI² (I: the current)
hence U = ½µ₀n²lA * (B/µ₀n)²
= B²/2µ₀ * Al (l: length)
u = U/lA = B²/2µ₀
derive the general potential at a point from a dipole
use superposition of the potentials from each charge
draw a dipole of separation d. r is the distance to the point from ½d.
r₁ is the distance from the top charge, r₂ is the distance from the bottom charge. θ is the angle between r and the verticle
r₁ = r - ½dcos(θ)
r₂= r + ½dcos(θ)
Δϕ = Q/4πε₀r₁ - Q/4πε₀r₂ = Q/4πε₀ [1/r₁ - 1/r₂]
taylor expand 1/r₁ and 1/r₂
1/r₁ = 1/[ r - ½dcos(θ)] = 1/r [1-d/2r cos(θ)]^-1
= 1/r[1 + d/2r cos(θ)]
hence 1/r₂ = 1/r[1 - d/2r cos(θ)]
hence 1/r₁ - 1/r₂ = 1/r[d/r cos(θ)]
ϕ ≈ Qdcos(θ)/4πε₀r²
derive the force on a current-carrying wire
F = q (v x B) (force on a particle)
hence the force on a wire:
F = N q(v x B) (N = nAL, the total number of particles)
since I = Anqv
F = AnLq(v x B) = L(I x B)
= BIL (if I and B are perpendicular)
derive the motional e.m.f equation and its direction for a square circuit being pulled through a B-field that’s into the page.
draw a rectangle, half of it is in the B-field. it is moving with velocity, v.
use F=q(v x B) (remember F is for a positive charge).
The electrons are flowing anticlockwise, hence the conventional current, I, is clockwise.
Magnetic flux, ϕ = ∫B⋅dA = Blx (l is the height of the circuit and x is the length of the circuit inside the B-field)
dϕ/dt = Bl *dx/dt = -Blv. (- because ϕ is decreasing)
dϕ/dt = -ε = -Blv
ε = Blv