karna (pk4534) – HW 07 – li – (59050)
1
This printout should have 32 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Consider a long wire and a rectangular current
loop.
A
B
C
D
I
1
4
b
a
I
2
Determine the magnitude and direction oF
the net magnetic Force exerted on the rectan
gular current loop due to the current
I
1
in the
long straight wire above the loop.
1.
5
F
=
μ
0
I
1
I
2
4
2
π
±
a
a
+
b
²
,down
2.
5
F
=
μ
0
I
1
I
2
2
π
±
ab
a
+
b
²
3.
5
F
=
μ
0
I
1
I
2
4
2
π
(
a

b
), right
4.
5
F
=
μ
0
I
1
I
2
4
2
π
(
a
+
b
), up
5.
5
F
=
μ
0
I
1
I
2
4
2
π
6.
5
F
=
μ
0
I
1
I
2
4
2
π
³
b
a
(
a
+
b
)
´
,up
correct
7.
5
F
=
μ
0
I
1
I
2
a
2
π4
(
b

a
), up
8.
5
F
=
μ
0
I
1
I
2
2
π
(
a

b
),leFt
Explanation:
To compute the net Force on the loop, we
need to consider the Forces on segments
AB
,
BC
,
CD
,and
DA
.The
net
Force on the loop
is the vector sum oF the Forces on the pieces oF
the loop. The magnetic Force on
AB
due to
the straight wire can be calculated by using
5
F
AB
=
I
2
µ
B
A
d5s
×
5
B.
In order to use this, we need to know the
magnitude and direction oF the magnetic feld
at each point on the wire loop. We can apply
the BiotSavart Law. The result oF this is that
the magnitude oF the magnetic feld due to the
straight wire is
B
=
μ
0
I
1
2
πr
,
and the direction oF the magnetic feld is given
by the right hand rule; the feld curls around
the straight wire with the feld coming out oF
the page above the wire and the feld going
into the page below the wire. We can now
fnd the Force on the segment
AB
;app
ly
ing
the right hand rule to fnd the direction oF the
cross product,
×
5
B
,weseethattheForcewill
be in the
up
direction. Since the wire along
the segment
AB
is straight and always at a
right angle to
5
B
,thecrossproducts
imp
l
ifes
to
Bds
.S
incethemagn
itudeo
Fthemagnet
ic
feld is constant along segment
AB
,i
tc
an
come out oF the integral which simplifes to
give us the result,
F
AB
=
I
2
4B
1
=
I
2
4
±
μ
0
I
1
2
πa
²
.
±ollowing the same argument, we see that the
Force on the segment
is
F
=
I
2
4
³
μ
0
I
1
2
π
(
a
+
b
)
´
,
and its direction is down.
This is because
the direction oF the current is now in in the
opposite direction along segment
!
We can do the use the same procedure For
segments
and
,butbecausethemag
netic feld decreases with distance From the
straight wire,
5
B
is changing along these seg
ments. This means that the integrals are not
as simple. Using the right hand rule, we see
that the Force on segment
is directed to
wards the right and the Force on segment
is directed towards the leFt. Because the two
segments oF wire are symmetrically placed,
their magnitudes will be equal. Since these