In a damped oscillating circuit the energy is dissipated In the resistor. The Q-factor Is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a ligt1y danced circuit the energy, U, in the circuit decreases according to the following equation. d u d t = − 2 β u , w h e r e β = R 2 L (b) Using the definition of the Q-factor as energy divided by the loss over the next cycle, prove that Q-factor of a lightly damped oscillator as defined in this problem is Q = U b e g i n △ U o n e c y c l e = 1 R L C c (Hint: For (b), to obtain Q, divide E at the beginning of one cycle by the change E over the next cycle.)
In a damped oscillating circuit the energy is dissipated In the resistor. The Q-factor Is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a ligt1y danced circuit the energy, U, in the circuit decreases according to the following equation. d u d t = − 2 β u , w h e r e β = R 2 L (b) Using the definition of the Q-factor as energy divided by the loss over the next cycle, prove that Q-factor of a lightly damped oscillator as defined in this problem is Q = U b e g i n △ U o n e c y c l e = 1 R L C c (Hint: For (b), to obtain Q, divide E at the beginning of one cycle by the change E over the next cycle.)
In a damped oscillating circuit the energy is dissipated In the resistor. The Q-factor Is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a ligt1y danced circuit the energy, U, in the circuit decreases according to the following equation.
d
u
d
t
=
−
2
β
u
,
w
h
e
r
e
β
=
R
2
L
(b) Using the definition of the Q-factor as energy divided by the loss over the next cycle, prove that Q-factor of a lightly damped oscillator as defined in this problem is
Q
=
U
b
e
g
i
n
△
U
o
n
e
c
y
c
l
e
=
1
R
L
C
c
(Hint: For (b), to obtain Q, divide E at the beginning of one
65. Although we usually write Newton's second law for one-dimensional
motion in the form F =ma, which holds when mass is constant,
d(mv)
a more fundamental version is F
=
.
Consider an object
dt
whose mass is changing, and use the product rule for derivatives to
show that Newton's law then takes the form F
dm
= ma + v
dt
If a proton is located on the x-axis in some coordinate system at x0 = -3.2 x 10-5 meters, what is the x-component of the Electric Field due to this proton at a position x = +3.2 x 10-5 meters and on the x axis as the y-axis is 0 giving a number of Newtons/Coulomb?
Consider a single square loop of wire of area A carrying a current I in a uniform magnetic field
of strength B. The field is pointing directly up the page in the plane of the page. The loop is
oriented so that the plane of the loop is perpendicular to the plane of the page (this means that the
normal vector for the loop is always in the plane of the page!). In the illustrations below the
magnetic field is shown in red and the current through the current loop is shown in blue. The
loop starts out in orientation (i) and rotates clockwise, through
orientations (ii) through (viii)
before returning to (i).
(i)
Ø I N - - I N -
(iii)
(iv)
(v)
(vii)
(viii)
a) [3 points] For each of the eight configurations, draw in the magnetic dipole moment vector
μ of the current loop and indicate whether the torque on the dipole due to the magnetic field
is clockwise (CW), counterclockwise (CCW), or zero. In which two orientations will the
loop experience the maximum magnitude of torque?
[Hint: Use the…
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