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PHY2049 Spring 2007
Exam 2 Solutions
1
Prof. Darin Acosta
Prof. Greg Stewart
March 26, 2007
Exam 2 Solutions
1. A current i travels in the direction of the arrows shown in the figure in three wire
segments.
The straight segments (upper left and upper right in the figure) both are lined
up with the center point,
P
, which is the center of the circular arc segment wire shown
which subtends (takes up) an angle of 120
o
.
The straight wires are both of length R; the
radius of the circular arc is also R.
What is the correct expression for the magnitude of
the total field produced by the current in all three wire segments at point P?
a.
μ
0
i/(6R)
b.
μ
0
i/(3R)
c.
μ
0
i/(2R)
d.
μ
0
i/(4R)
e.
μ
0
i/(12R)
Solution:
The current in the straight wires that would go thru center point P if extended
contributes zero to the field (as discussed in lecture for chap. 29).
The field at the center
of a circular arc of wire (eq. 299) is B=
μ
0
i
φ
/(4
π
R),
where
φ
is the angle taken up by the
circular arc.
120
o
is 1/3 of 360
o
, where 360
o
is 2
π
radians, so 120
o
=(1/3) 2
π
radians, so
B=
μ
0
i
φ
/(4
π
R)= =
μ
0
i(2/3)
π
/(4
π
R)=
μ
0
i/(6R).
2.
Consider the toroid shown in the figure.
Wire is wrapped around the core in a
doughnut geometry, current i=10 A in the wire.
As shown, the wire carries current
out
of
the page (this direction of the wire is shown as small 'o'), around the toroid, and back
into
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Exam 2 Solutions
2
the page (denoted as 'x').
There are 40 windings or turns on this toroid.
Three circular
Amperian loops – the smallest one (#1), the middlesized one between the inner and outer
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