Physics 133: tutorial week 5
Alternating currents, magnetic fields and transformers.
61. In the wires connecting an electric clock to a wall socket, how many times a day
does the alternating current reverse its direction if run off South African mains?
(8.64 × 106 )
The current reverses direction twice per cycle and there are 50 cycles per second. The number
of reversals per day is therefore 2 × 50 times the number of seconds in a day:
Reversals per day = (2 × 50) × 24 × 60 × 60 = 8.64 × 106 .
62. Calculate the peak current passing through a 60 W light bulb connected to a 240 V
ac supply.
(0.354 A)
60
P
=
= 0.25 A .
Irms =
V
240
√ rms
√
I0 = 2 × Irms = 2 × 0.25 = 0.354 A .
63. An ac voltage, whose peak value is 90 V, is across a 35 W resistor. What is the value
of the peak and rms currents in the resistor?
(2.57 A; 1.82 A)
I0 =
Irms
90
V0
=
= 2.57 A .
R
35
2.57
I0
= √ = √ = 1.82 A .
2
2
64. Suppose that a current is given by the equation: I = 2.5 sin(120t) (with I in amps
and t in seconds).
(a) What is the frequency?
(b) What is the rms value of the current?
(c) If this current is through a 50 W resistor, what is the equation that describes the
voltage as a function of time?
(19.1 Hz ; 1.77 A ; 125 sin(120t) volts)
(a) 2πf = 120 Hz
f = 19.1 Hz .
2.5
(b) I0 = 2.5 V ∴ Irms = √ = 1.77 A .
2
(c) V0 = I0 R = 2.5 × 50 = 125 V ∴ V = 125 sin(120t) volts .
∴
65. A wire 1.0 m long carries a current of 10 A and makes an angle of 30◦ with a uniform
magnetic field with B = 1.5 T. Calculate the magnitude and direction of the force on
the wire.
(7.5 N , perpendicular to both the wire and to B)
F = BIℓ sin θ = 1.5 × 10 × 1.0 × sin 30◦ = 7.5 N .
30◦
Using the right hand push rule for the configuration depicted on the right, the force is into the page, at right
angles to both the field and the wire.
66. (a) Calculate the force per unit length on a wire carrying a current of 0.5 A when
perpendicular to a 4.0 T magnetic field.
(b) What if the angle between the wire and the field is 45◦ ?
(2.0 N m−1 ; 1.41 N m−1 )
(a) The wire is perpendicular to the field, therefore θ = 90° and sin θ = 1 hence F = BIℓ .
The force per unit length is:
F
= BI = 4.0 × 0.5 = 2.0 N m−1 .
ℓ
F
= BI sin θ = 4.0 × 0.5 × sin 45° = 1.41 N m−1 .
(b)
ℓ
67. The force on a wire carrying 20 A is a maximum of 3.6 N when placed between the
pole faces of a magnet. If the wire is 15 cm long, what is the approximate magnitude
of B?
(1.2 T)
The force is a maximum when the angle the wire makes with the field is 90° . Hence
3.6
F
=
= 1.2 T .
B =
Iℓ
20 × 0.15
68. How much current is flowing in a wire 2.0 m long if the maximum force on it is 0.7 N
when placed in a uniform 0.03 T field?
(11.7 A)
Since force is a maximum, the angle between the wire and the field is 90° . Hence
0.7
F
=
= 11.7 A .
I =
Bℓ
0.03 × 2.0
69. Calculate the magnetic force on a 140 m length of wire stretched between two towers
carrying a 200 A current. The earth’s magnetic field of 5 × 10−5 T makes an angle of
60◦ with the wire.
(1.212 N)
F = BIℓ sin θ = 5 × 10−5 × 200 × 140 × sin 60° = 1.212 N .
70. Calculate the magnitude of the magnetic field B, 32 cm from a long, straight wire
carrying 8.0 A.
(5.0 × 10−6 T)
B = k
I
8.0
= 2 × 10−7 ×
= 5.0 × 10−6 T .
r
0.32
71. How much current does a wire carry if B = 0.03 T at a point 12 cm from it? (18000 A)
I =
0.03 × 0.12
Br
= 18000 A .
=
k
2 × 10−7
72. How far from a long, straight wire carrying 10 A will the magnetic field be 1.0 ×
10−2 T?
(0.20 mm)
r =
kI
2 × 10−7 × 10
= 2 × 10−4 m = 0.20 mm .
=
B
1.0 × 10−2
73. Determine the magnetic field midway between two long, straight wires 10 cm apart
if one carries 10 A and the other 8.0 A and these currents are
(a) in the same direction, and
(b) in opposite directions.
(8 × 10−6 T ; 72 × 10−6 T)
(a) Midway between the wires, the field due to each wire is perpendicular to the line joining
the wires. When the current in the wires is in the same direction, the fields are in opposite
directions. Hence 10
I2
8.0
I1
= 2 × 10−7
= 8.0 × 10−6 T .
B = B1 − B2 = k
−
−
r
r
0.05 0.05
(b) When the wires carry current in the opposite direction, the fields are in the same direction,
hence
I1
10
I2
8.0
−7
B = B1 + B2 = k
= 2 × 10
= 72 × 10−6 T .
+
+
r
r
0.05 0.05
74. Two long, fixed vertical wires A and B are 50 cm apart and carry currents of 100 A
and 60 A respectively in opposite directions. Determine
(a) the force per unit length exerted by wire A on wire B,
(b) the magnetic field at P 100 cm from A and 150 cm from B, and
(c) the position of any neutral point in a horizontal plane.
(2.4 × 10−3 N m−1 away from A; 1.2 × 10−5 T; 75 cm from B and 125 cm from A)
F
kIA IB
2 × 10−7 × 100 × 60
=
=
= 2.4 × 10−3 N m−1 (away from A).
ℓ
L
0.50
(b) The point 100 cm from A and 150 cm from B lies on the line joining A and B since the wires
are 50 cm apart. The field due to both wires at this point is therefore perpendicular to
the line joining the wires. Furthermore, The fields here are in opposite directions. Hence
100
IB
60
IA
= 1.2 × 10−5 T .
= 2 × 10−7
−
B = BA − BB = k
−
rA
rB
1.0
1.5
(c) There will be a neutral point where BA = BB , a distance x from A so that rA = x and
rB = x − 0.5 (we take x − 0.5 instead of x + 0.5 since the neutral point lies closer to B than
to A because the field due to A is stronger than the field due to B). Hence
IA
IB
100
60
k
= k
gives
=
so that x = 1.25 m .
rA
rB
x
x − 0.5
(a)
75. Two long, parallel wires 12 cm apart carry 15 A currents in the same direction. Determine the magnitude of the magnetic field B at a point 10 cm from one wire and
20 cm from the other. (Hint: make a drawing in a plane containing the field lines
and recall the rules for vector addition.)
(4.4 × 10−5 T)
R
BR
BB
BA
B′
A′
P
cm
10
A
20 c
m
12 cm
B
P
θ
A
B
BA and BB denote the magnetic fields at point P: they are tangential to the circles centred at
A and B respectively. BR is the resultant.
IA
15
BA = k
= 2 × 10−7 ×
= 3.0 × 10−5 T .
rA
0.10
15
IB
= 2 × 10−7 ×
= 1.5 × 10−5 T , and
BB = k
rB
0.20
BR2 = BA2 + BB2 − 2BA BB cos φ .
Here φ = ∠PA′ R = 180° − θ , since PA′ RB′ is a parallelogram with ∠P = ∠R = θ . In △PAB ,
122 = 102 + 202 − 2 × 10 × 20 × cos θ ,
which gives
cos θ = − cos φ = 0.89 .
Finally,
1
BR = (3.02 + 1.52 + 2 × 3.0 × 1.5 × 0.89) 2 × 10−5 = 4.39 × 10−5 T.
76. Calculate the magnitude and direction of the force between two parallel wires 85 m
long and 30 cm apart, each carrying 60 A in the same direction. (0.204 N attractive)
F
kI1 I2
=
ℓ
L
∴
F =
2 × 10−7 × 60 × 60 × 85
= 0.204 N .
0.3
77. A vertical, straight wire carrying 5 A exerts an attractive force per unit length 6 ×
10−4 N m−1 on a second, parallel wire 8.0 cm away. What current (magnitude and
direction) flows in the second wire?
(48 A)
kI1 I2
F
=
ℓ
L
∴
I2 =
0.08
F
L
= 6 × 10−4 ×
×
= 48 A .
ℓ
kI1
2 × 10−7 × 5
78. A transformer changes 12 V to 18000 V and there are 6000 turns in the secondary.
How many turns are there in the primary?
(4)
np = ns ×
Vp
12
= 6000 ×
= 4.
Vs
18000
79. A transformer has 145 turns in the primary and 55 in the secondary. What kind of
transformer is this, and by what factor does it change the voltage?
(step down; 0.379)
This is a step-down transformer, with
55
ns
=
= 0.379 .
np
145
80. A step-up transformer increases 30 V to 120 V. What is the current in the secondary
as compared to the primary?
(IS = IP /4)
np
30
Is
=
=
Ip
ns
120
∴
Is =
1
4 Ip
81. Describe a transformer that could be used to light a 6 V bulb from a 240 V, 50 Hz
source.
(step down; turns ratio of 40)
We require a step-down transformer with
Vs
6
1
ns
=
.
=
=
Vp
240
40
np
82. A transformer has 1500 primary turns and 120 secondary turns. The input voltage
is 240 V and the output current is 8.0 A. What is the secondary voltage and primary
current?
(19.2 V; 0.64 A)
1500
240
Is
8.0
np
Vp
=
=
=
=
=
ns
120
Vs
Vs
Ip
Ip
240×120
8.0×120
∴ Vs =
= 19.2 V and Ip =
= 0.64 A .
1500
1500