Discussion Question 10A
P212, Week 10
Faraday's Law: Moving Loop
A one-turn current loop in the shape of an equilateral triangle with sides a lies in the xy plane
and moves with velocity v in the +x direction. (A magic external force keeps the velocity
constant ☺) The loop has a net resistance R . It passes through a region of constant, spatially
uniform magnetic field B that points in the -z direction (into the page) and extends from x = 0 to
x=d.
y
a
a
a
v
x
0
x
d
We will specify the loop’s position using the x-coordinate of the triangle’s tip (as shown in the
figure). Now, before we do any calculations, let’s think physically …
(a) During what part(s) of the loop’s trip (i.e. range(s) in x-position) will there be a nonzero current induced in the loop?
Remember, induced EMF only occurs when the magnetic flux through a loop is changing.
entering: x = 0 to √3*a/2
exiting: x = d to d + √3*a/2
(b) Sketch the induced current I as a function of x over the full range of motion shown on
the figure. Let positive I indicate current flow in the clockwise direction.
In computing the flux, we take the normal to be in the
I
direction of the positive current direction according to
the righthand rule. This direction would be into the page
of the paper and hence the flux is positive Φ = BA
dΦ
t
.
where A is the area of the loop in the field. E = −
dt
Hence as the triangle enters the loop Φ is increasing and E < 0 →
I < 0. As the triangle exits the loop Φ is decreasing and hence
E > 0 → I > 0. The rate of change of Φ is proportional to the
velocity times the base of the triangle and hence you get the
ramped shape.
y
a
a
v
a
x
0
x
d
(c) Derive an analytic expression (i.e. no numbers ☺) for the current I induced in the loop
as a function of the loop's x-position. This expression should be valid from the moment the tip
of the triangle enters the magnetic field until the base of the triangle enters the field.
(i) To obtain the induced current, you first need an expression for the magnetic flux
ΦΒ through the loop as a function of the loop's position x. The B-field part of this expression
is easy ... the area part requires some thought. (Hint: what is the height of an equilateral
triangle of side a?)
(ii) Now take the time derivative of your expression, to get an EMF.
(iii) The final step is to turn your induced EMF into a current .
The key insight is to realize that the part of the loop covered
by the B field is still an equilateral +. As the figure shows
the height of this triangle is x and the base is 2 x tan 300.
300
x
1
x2
2
0
Hence the area is A= base × height = x tan 30 =
2
3
2
x
The flux is then Φ = ΒΑ = Β
3
2
dΦ
2B
B dx
B dx 2 dx
=−
=−
=−
ε =−
xv
dt
3 dt
3 dx dt
3
E
2 Bxv
I = =−
R
3R