Download A circuit with “free

A circuit with “free-falling” parts
A circuit of rectangular shape and resistance R is place vertically in the gravity field g. The two
horizontal sides have length ℓ, mass M and can move freely in vertical direction keeping the electrical
contact. The system is placed in an uniform and constant magnetic field B perpendicular to the
plane of the circuit. (The variation of the resistance with the position of the moving sides and the
self-induction of the circuit are assumed to be negligible).
R/ 2
R/ 2
B
g
v1
R/ 2
R/ 2
v
v2
l
l
B
B
Firstly, assume that the upper side is kept fixed, while the lower side starts its motion at t = 0.
Let v = v(t) its velocity at the time t, and v(0) = 0.
a) Find the equation of motion for v(t) and give its solution showing that asymptotically the velocity
assumes a constant value vd .
b) When v(t) = vd find the power dissipated in the circuit due to Joule heating and the mechanical
power due to the gravity force.
Now consider the case in which both moving sides are freed for t > 0, with the initial conditions
v1 (0) = v0 6= 0 and v2 (0) = 0.
c) Find the equations of motion for the velocities of the two mobile sides v1 (t) and v2 (t) and its
solution. Discuss the asymptotic behavior of v1 (t), v2 (t) and of the current in the circuit I(t).
1
Solution
a) Let us fix the y axis oriented upwards and with the position of the upper side at y = 0. The
current I in the circuit is given by
dΦ(B)
d(−y)
= −Bℓ
= +Bℓv.
(1)
dt
dt
The velocity has negative sign and the current circulates clockwise, in agreement with Lenz’s law.
The force on the moving side is F = −BℓI ŷ and it is directed oppositely to g. Thus
RI = E = −
(Bℓ)2
dv
= −M g − BℓI = −M g −
v.
dt
R
The solution of (2) with v(0) = 0 is
MR
M Rg
τ=
v(t) = vd 1 − e−t/τ ,
,
vd = −gτ = −
.
2
(Bℓ)
(Bℓ)2
M
For t → ∞, the side falls with constant drift velocity vd .
b) When v(t) = vd the power dissipated by Joule heating is
2
(Bℓv)2
Mg
2
Pd = RI =
R.
=
R
Bℓ
(2)
(3)
(4)
The mechanical power is
Pm = M g · v = M g
M gR
= Pd ,
(Bℓ)2
(5)
in agreement with energy conservation.
c) The current is now given by
dΦ(B)
d
= −Bℓ (y1 − y2 ) = +Bℓ(v2 − v1 ).
(6)
dt
dt
On the two mobile sides we have F1 = BℓI ŷ and F2 = −BℓI ŷ, respectively. The equations of motion
are
(Bℓ)2
dv2
(Bℓ)2
dv1
= −M g +
(v2 − v1 ),
M
= −M g −
(v2 − v1 ).
(7)
M
dt
R
dt
R
By taking the sum and the difference of (7) we obtain
RI = E = −
d
1
(v2 − v1 ) = − (v2 − v1 ),
dt
τ
which have the solutions
v2 − v1 = v0 e−t/τ ,
d
(v2 + v1 ) = −2g,
dt
v2 + v1 = v0 − 2gt,
(8)
(9)
from which we obtain:
v1,2 =
v0
(1 ∓ e−t/τ ) − gt.
2
(10)
For t ≫ τ
v1 ≃ v2 ≃
v0
− gt.
2
The current I(t) = 0 as v1 = v2 .
2
(11)