Download Quiz 6 - Rutgers Physics

Solutions to Quiz 6
April 28, 2010
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Magnetic Induction
Two parallel metal rails are connected at one end (see figure). At a distance
x from the end, there is a movable metal rod of length L that connects the
rails and completes a loop. A uniform magnetic field B points out of the
page.
1.1
Solution
a. Find the magnetic flux through
the loop.
R
~ · d~a = BLx
The magnetic flux is ΦB = B
b. Is there an induced current in the rod? Why or why not?
There is no induced current as the magnetic flux through the loop is not
changing.
c. You now give the movable rod a nudge to the left. It acquires a velocity
v. At that instant, find the rate of change of magnetic flux (dφ/dt) through
the loop. Is the magnitude of the flux increasing or decreasing?
The rate of change of flux is
dΦB
dx
= BL
= BLv
dt
dt
(1)
and the magnitude is increasing because the area is increasing creating a
greater flux through the loop.
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d. If the rod has resistance R and the rails have negligible resistance, find
the magnitude and direction of the current i through the loop. Your answer
should be expressible in terms of B, L, v and R.
The induced current is given by i = /R where R is the resistance of the
loop and is the induced emf. The induced emf is given by Faraday’s Law:
=
dφ
= BLv
dt
(2)
and so the current is i = BLv/R and is clockwise (so that the magnetic field
that the current creates opposes the change in magnetix flux, this is Lenz’s
Law).
e. As a result of this current, a new force acts on the rod. Find the
magnitude and direction of the force on the rod. Your answer should be
expressible in terms of B, L, v and R.
The new force that acts on the rod is the magnetic force on a current,
given by
~ ×B
~
F~B = iL
(3)
which has a magnitude FB = B 2 L2 v/R and a direction to the right.
f. Describe the motion of the rod for all times. Assume (as always) that
there is no friction between the rod and the rails.
Since the force opposes the motion of the rod, the rod will decelerate. It
will eventually come to a stop because when v = 0, FB = 0.
g. Solve for x(t) i.e. find the position of the rod as a function of time.
The force equation is
F =m
B 2 L2
dv
=−
v
dt
R
(4)
where the negative sign is because the force opposes the velocity of the rod.
This can be re-written as
dv
= −Cv
dt
(5)
where C = B 2 L2 /R is a constant with units of sec−1 . Thus, 1/C gives us a
time (also called the time constant) which will give us an idea of how long
this rod takes to stop.
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This equation can be re-arranged and integrated
Z v
v0
Z t
dv
= −C
dt
v
0
(6)
giving ln(v/v0 ) = −Ct, or taking exponential of both sides,
v = v0 e−Ct
(7)
In words, the speed dies out exponentially. If you wait 1/C seconds, the
velocity decreases by a factor of e. As t → ∞, v → 0.
We can then integrate this equation once more to find the position as a
function of time
dx
= v0 e−Ct
(8)
dt
or
Z x
Z t
dx = v0
e−Ct dt
(9)
0
0
Solving and rearranging,
x=
v0 1 − e−Ct
C
(10)
If you plot this function, you’ll see that the position grows towards vC0
exponentially quickly, but it never crosses this distance, as expected from
the previous part. As t → ∞, x → v0 /C, and this is where the rod stops.
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