The magnetic field at A due to
the upward current is
B
up,A
=
µ
0
(2
I
0
)
2
π
(
r/
2)
=
2
µ
0
I
0
π r
.
The right-hand rule tells us the direction is
into the paper.
Due to the fact that A is
the same distance from both wires, the total
magnetic field at A is
B
A
=
2
µ
0
I
0
π r
+
µ
0
I
0
π r
=
3
µ
0
I
0
π r
.
Now, the field at B due to the upward current
is
B
up,B
=
2
µ
0
I
0
2
π
(3
r/
2)
=
2
µ
0
I
0
3
π r
again into the paper, while the downward
current gives
B
down,B
=
µ
0
I
0
2
π
(
r/
2)
=
µ
0
I
0
π r
out of the paper. So at B, the net field out of
the paper is:
B
B
=
B
down,B
−
B
up,B
=
µ
0
I
0
π r
parenleftbigg
1
−
2
3
parenrightbigg
=
µ
0
I
0
3
π r
.

mittal (im5936) – Magnetic Force and Field HW – yeazell – (58010)
9
Comparing their magnitudes, we find
B
A
B
B
=
3
µ
0
I
0
π r
µ
0
I
3
π r
=
9
.
020
10.0points
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.
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,