MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2007
Problem Set 7 Solutions
Problem 1:
Quickies…
a)
Two semicircular arcs have radii
R
2
and
R
1
, carry
current
i
, and share the same center of curvature C.
What is the magnitude of the net magnetic field at C?
The inner semi-circle makes a field into the page, the outer one out of the page.
The
inner is closer and hence stronger, and hence the net field is into the page.
In class you
calculated the magnetic field from a semi-circle of radius
R
to be
0
4
B
i
R
μ
=
.
So:
0
1
1
into the page
4
i
R
R
μ
⎛
⎞
=
−
⎜
⎟
B
G
1
2
⎝
⎠
le, what is
θ
?
b)
A wire with current
i
is shown at left. Two semi-infinite
straight sections, both tangent to the same circle with
radius
R
, are connected by a circular arc that has a
central angle
θ
and runs along the circumference of the
circle. The connecting arc and the two straight sections
all lie in the same plane. If
B
= 0 at the center of the
circ
The straight portions both make a field out of the page at the center of the circle while the
arc makes one into the page.
These must be equal so that the fields cancel.
Two semi-
infinite lines together make an infinite line, and we calculated (using Ampere’s law) that
the field from an infinite wire is
0
2
B
i
R
μ
=
π
.
The arc is just a fraction of a circle so it
creates a fraction of the field that a whole circle does at its center:
(
)(
)
0
2
2
B
i
R
μ
=
θ
π
.
For these to be equal we must have
0
0
2
4
2 radians
B
i
R
i
R
μ
μ
=
π
=
θ
π
⇒
θ =
c)
The figure at left shows two closed paths wrapped around
two conducting loops carrying currents
i
1
and
i
2
. What is
the value of the integral for
(a)
path 1 and
(b)
path 2?
To do this you have to use the right hand rule to check whether the currents are positive
or negative relative to the path.
On path 1
i
1
penetrates in the negative direction while
i
2
penetrates in the positive direction, so
(
)
2
1
o
d
i
i
⋅
= μ
−
∫
B
s
G
G
v
.
On path 2 i
1
penetrates twice in the negative direction and i
2
once in the negative
direction so
(
)
1
2
2
o
d
i
i
⋅
= −μ
+
∫
B
s
G
G
v
PS07 Solutions-1

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