B w l 2 μ b sin90º w 2 μ b l 2 ia b l 2 il 2 b l 2

B
!"
W
L
2
⎛
⎝
⎜
⎞
⎠
⎟
= μ B sin90º ⇒ W =
2 μ B
L
=
2 IA
( )
B
L
=
2 IL 2 B
L
= 2 ILB
W = 2 3.6 A
( )
0.3 m
( )
0.4 T
( )
= 0.864 N
shared via CourseHero.com
B =
μ
o
I
2 r
±
+
−
R
B
!"
= ?
R
6
This study resource was
shared via CourseHero.com

Now we need only to find the four currents. The current through segments B and C equal the current through the
horizontal resistor, and since the potential difference across that resistor is the voltage of the battery, we have simply:
The current through conductors A and D are the same as the current from the battery, and for this current we need to
compute the equivalent resistance. The resistors are in parallel, so the equivalent resistance is easy to find, and with it
the two currents:
Putting it all together:
B
A
=
1
4
μ
o
I
A
2 a
⎛
⎝
⎜
⎞
⎠
⎟
out of page
!
( )
B
B
=
1
4
μ
o
I
B
2 a
⎛
⎝
⎜
⎞
⎠
⎟
out of page
!
( )
B
C
=
1
4
μ
o
I
C
2 2 a
( )
⎛
⎝
⎜
⎞
⎠
⎟
into page
!
( )
B
D
=
1
4
μ
o
I
D
2 2 a
( )
⎛
⎝
⎜
⎞
⎠
⎟
into page
!
( )
⎫
⎬
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⇒ B
tot
=
μ
o
8 a
⎛
⎝
⎜
⎞
⎠
⎟
I
A
+ I
B
−
1
2
I
C
−
1
2
I
D
⎡
⎣
⎢
⎤
⎦
⎥
out of page
!
( )
I
B
= I
C
=
ε
R
1
R
eq
=
1
R
+
1
R
=
2
R
⇒ R
eq
=
R
2
⇒ I
A
= I
D
=
ε
R
eq
=
2
ε
R
B
tot
=
μ
o
8 a
⎛
⎝
⎜
⎞
⎠
⎟
2
ε
R
+
ε
R
−
1
2
ε
R
⎛
⎝
⎜
⎞
⎠
⎟
−
1
2
2
ε
R
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥
out of page
!
( )
=
3 μ
o
ε
16 aR
out of page
!
( )
7
This study resource was
shared via CourseHero.com
Powered by TCPDF ()