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karna (pk4534) – HW 07 – li – (59050) 1
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001 10.0 points
Consider a long wire and a rectangular current
loop.
A
B
C
D
I
1
ℓ
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.
vector
F =
μ
0
I
1
I
2
ℓ
2 π
parenleftbigg
a
a + b
parenrightbigg
, down
2.
vector
F =
μ
0
I
1
I
2
2 π
parenleftbigg
a b
a + b
parenrightbigg
, down
3.
vector
F =
μ
0
I
1
I
2
ℓ
2 π
( a − b ), right
4.
vector
F =
μ
0
I
1
I
2
ℓ
2 π
( a + b ), up
5.
vector
F =
μ
0
I
1
I
2
ℓ
2 π
a b , down
6.
vector
F =
μ
0
I
1
I
2
ℓ
2 π
bracketleftbigg
b
a ( a + b )
bracketrightbigg
, up correct
7.
vector
F =
μ
0
I
1
I
2
a
2 π ℓ
( b − a ), up
8.
vector
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
vector
F
AB
= I
2
integraldisplay
B
A
dvectors ×
vector
B .
In order to use this, we need to know the
magnitude and direction of the magnetic field
at each point on the wire loop. We can apply
the Biot-Savart Law. The result of this is that
the magnitude of the magnetic field due to the
straight wire is
B =
μ
0
I
1
2 π r
,
and the direction of the magnetic field is given
by the right hand rule; the field curls around
the straight wire with the field coming out of
the page above the wire and the field going
into the page below the wire. We can now
find the force on the segment AB ; applying
the right hand rule to find the direction of the
cross product, dvectors ×
vector
B , we see that the force will
be in the up direction. Since the wire along
the segment AB is straight and always at a
right angle to
vector
B , the cross product simplifies
to B ds . Since the magnitude of the magnetic
field is constant along segment AB , it can
come out of the integral which simplifies to
give us the result,
F
AB
= I
2
ℓ B
1
= I
2
ℓ
parenleftbigg
μ
0
I
1
2 π a
parenrightbigg
.
Following the same argument, we see that the
force on the segment CD is
F
CD
= I
2
ℓ
bracketleftbigg
μ
0
I
1
2 π ( a + b )
bracketrightbigg
,
and its direction is down. This is because
the direction of the current is now in in the
opposite direction along segment CD !
We can do the use the same procedure for
segments BC and DA , but because the mag-
netic field decreases with distance from the
straight wire,
vector
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 BC is directed to-
wards the right and the force on segment DA
is directed towards the left. Because the two
segments of wire are symmetrically placed,
their magnitudes will be equal. Since these

karna (pk4534) – HW 07 – li – (59050) 2