HW #6 Solutions
19.38
The magnitude of the field due to a single wire is given by:
µoI (4π x 10-7 T m/A)(5.00 A)
1.00 x 10-6 T m
B=
=
=
.
r
2πr
2πr
Call the wire on the left wire 1 and the wire on the right wire 2.
(a) At a point midway between the wires, r1 = r2 = 5.00 cm. Thus,:
1.00 x 10-6 T m
= 2.00 x 10-5 T. Using the right hand rule, we see that both
0.0500 m
fields are in the same direction (into the page), which gives us a resultant field of:
Bnet = B1 + B2 = 4.00 x 10-5 T
(directed into page).
B1 = B2 =
(b)
At the point P1: B1 =
1.00 x 10-6 T m
= 5.00 x 10-6 T (into the page), and B2 =
0.200 m
1.00 x 10-6 T m
= 10.0 x 10-6 T (out of the page).
0.100 m
Thus, Bnet = 5.00 x 10-6 T (out of page).
(c)
19.39
1.00 x 10-6 T m
= 5.00 x 10-6 T (out of the page), and
0.200 m
1.00 x 10-6 T m
= 3.33 x 10-6 T (into the page),
B2 =
0.300 m
for Bnet = 1.67 x 10-6 T (out of page).
At P2,: B1 =
Call the wire on the left 1 and the wire on the right 2.
(b) (See the sketch) The distance R is
B1
θ = 45 °
R = (0.2 m)2 + (0.2 m)2 = 0.283 m.
The field produced by wire 2 at P2 is:
µoI (4π x 10-7 T m/A)(5 A)
=
B2 =
2πr
2π(0.2 m)
= 5.00 x 10-6 T
and the field from wire 1 at point P2 is
µoI (4π x 10-7 T m/A)(3 A)
=
, or
B1 =
2πR
2π(0.283 m)
B1 = 2.12 x 10-6 T. The angle θ shown in the sketch is 45 °.
B2
R
3A •
Wire 1
0.2 m
P2
0.2 m
• 5A
Wire 2
Thus, the fields can be
broken into their x and y components. These resultant components are: Bx = -6.5 x 10-6 T,
By = 1.5 x 10-6 T. The Pythagorean theorem gives a resultant field of:B = 6.67 x 10-6 T at
77 ° to the left of the vertical.
19.43
In order for the system to be in equilibrium, the magnetic force per unit length on the top wire
must be equal to its weight per unit length.
µoI1I2
F
F
Thus,
= weight per unit length, where:
=
.
L
L
2πd
(30.0 A)(60.0 A)
We have, (2 x 10-7 T m/A) 
d

 = 0.080 N/m, and find
d = 4.50 x 10-3 m = 4.50 mm.
19.47
(a)
The magnetic force provides the centripetal acceleration
mv2
mv
(9.11 x 10-31 kg)(104 m/s)
qvB =
, or B =
=
= 2.8 x10-6 T
qr (1.60 x 10-19 C)(2.0 x 10-2 m)
R
B
2.8 x 10-6 T
(b) From B = µonI:
I=
=
= 8.9 x 10-4 A =
µon
(4π x 10-7 Tm/S)(2500 turns/m)
0.89 mA.
19.49
Call the wire along the y-axis #1 and the wire along the x-axis #2.
(b) We use the same designation for discussing the wires as above. The field due to wire 1 is:
µoI1 (4π x 10-7 T m/A)(3.00 A)
+z
B1 =
=
= 2.00 x 10-6 T
2πr1
2π(0.300 m)
B
The sketch indicates the direction of the field.
The field due to wire 2 is:
µoI2 (4π x 10-7 T m/A)(5.00 A)
B2 =
=
= 3.33 x 10-6 T
2πr2
2π(0.300 m)
(For direction, see sketch.
The net field is found from the Pythagorean theorem to be
Bnet = 3.89 x 10-6 T.
The angle, as seen from above the wires (second
sketch) is:
B1
tanθ =
B2 = 0.6 and θ = 31° (CCW from -y direction).
1
2
0.3 m
B
+y
I1 = 3 A
+x
I 2= 5 A
B1
θ
B2
Bnet
2. Consider the copper tube to be a large set of rings stacked one on top of the other. As
the magnet falls toward or falls away from each ring, a current is induced in the ring.
Thus, there is a current in the copper tube around its circumference.
20.3
Φ = BAcosθ Thus,
(a) Φ = (5.00 x 10-5 T)(20.0 x 10-4 m2)cos(0°) = 1.00 x 10-7 T m2.
(b) Φ = (5.00 x 10-5 T)(20.0 x 10-4 m2)cos(30°) = 8.66 x 10-8 T m2.
(c) Φ = (5.00 x 10-5 T)(20.0 x 10-4 m2)cos(90°) = 0.
y
20.7 Φ = BAcosθ = (component of B perpendicular to surface)A
(a) Φshaded side = BxA = (5.0 T)(2.5 x 10-2 m)2
= 3.1 x 10-3 T m2.
(b) Magnetic field lines have no beginning and no end. Thus, with a
uniform magnetic field, any field line that emerges from the cube
through one face entered the cube through the opposite face.
Therefore, the net flux (number of field lines) emerging from the
cube is zero. Φtotal = 0.
20.11
l
l
z
To produce 0.10 A in an 8.0 Ω coil, the induced emf must be
ε
= IR = (0.10 A)(8.0 Ω) = 0.80 V.
∆(BA)
∆B
∆Φ
∆Φ
where,
=
=A
.
Also, ε = N
∆t
∆t
∆t
∆t
∆B
∆B
Thus, 0.80 V = (75)
(0.050 m)(0.080 m) , and
= 2.7 T/s.
∆t
∆t
20.14
(a)
(b)
The magnetic field set up inside the solenoid is:
100 turns
B = µonI = (4π x 10-7 Tm/A)
(3.00 A) = 1.88 x 10-3 T, and
0.200 m
Φi = BA = (1.88 x 10-3 T)(10-2 m)2 = 1.88 x 10-7 T m2.
When the current is reduced to zero, Φf = 0.
∆Φ = 1.88 x 10-7 T m2, and
1.88 x 10-7 T m2
∆Φ
>=N
=
= 6.28 x 10-8 V.
3.00 s
∆t
Therefore,
<ε
B
l
x