Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf
Electromagnetics –
lecture summary
Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf
, Electromagnetics.pdf Electromagnetics.pdf Electromagnetics.pdf
C
"Magnetic Flux": H = B / μ = (I * Aф) / 2ℼr
Ф = BA
Given:
I =5A
A=3m
Ф = 30 mWb
I_new = 4 A
Ф_new / I_new = Ф_old / I_old
Ф_new = [ (30 mWb) / (5 A) ] * (4 A)
Ф_new =24 mWb
D
"Electrostatics" : F2 = [ (Q1 Q2) / (4 ℼ r^2 ε) ] A_12
"The net force on center = 0"
Zero net: F_2to3 = F_5to3
[ (Q2 Q3) / (4 ℼ [2-x]^2 ε) ] = [ (Q3 Q5) / (4 ℼ (x)^2 ε) ]
Q2 / [2-x]^2 = Q5 / x^2
...
2x^2 = 20 - 20x + 5x^2
Use the quadratic equation
X = 1.23 or 5.44, but the total was 2, so we know x = 1.23.
Electromagnetics.pdf Electromagnetics.pdf Electromagnetics.pdf
Electromagnetics –
lecture summary
Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf Electromagnetics – lecture summary.pdf
, Electromagnetics.pdf Electromagnetics.pdf Electromagnetics.pdf
C
"Magnetic Flux": H = B / μ = (I * Aф) / 2ℼr
Ф = BA
Given:
I =5A
A=3m
Ф = 30 mWb
I_new = 4 A
Ф_new / I_new = Ф_old / I_old
Ф_new = [ (30 mWb) / (5 A) ] * (4 A)
Ф_new =24 mWb
D
"Electrostatics" : F2 = [ (Q1 Q2) / (4 ℼ r^2 ε) ] A_12
"The net force on center = 0"
Zero net: F_2to3 = F_5to3
[ (Q2 Q3) / (4 ℼ [2-x]^2 ε) ] = [ (Q3 Q5) / (4 ℼ (x)^2 ε) ]
Q2 / [2-x]^2 = Q5 / x^2
...
2x^2 = 20 - 20x + 5x^2
Use the quadratic equation
X = 1.23 or 5.44, but the total was 2, so we know x = 1.23.
Electromagnetics.pdf Electromagnetics.pdf Electromagnetics.pdf
