Problem 1. (a) Find the direction of the force on a proton (a positively charged particle)
moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the
moving particle is an electron.
Solution: (a)
For a positively charged particle, the direction of the force is
that predicted by the right hand rule. These are:
(a’)
in plane ofpage and to left
(b’)
(c’)
outofthe page
(d’)
into the page
in plane ofpage and tow ard the top
(e’)
into the page
(f’)
outofthe page
(b) For a negatively charged particle, the direction of the force is exactly
opposite what the right hand rule predicts for positive charges. Thus, the
answers for part (b) are reversed from those given in part(a) .
Problem 2. Find the direction of the magnetic field acting on the positively charged
particle moving in the various situations shown in the figure, if the direction of the magnetic
force acting on it is as indicated.
Solution: Since the particle is positively charged, use the right hand rule. In this
case, start with the thumb of the right hand in the direction of v and the palm
facing the direction of F. The fingers will point in the direction of B. The
results are
(a)
into the page
(b)
tow ard the right
(c)
tow ard bottom ofpage
Problem 3. Determine the initial direction of the deflection of charged particles as they
enter the magnetic fields shown in the figure.
Solution: Hold the right hand with the thumb in the direction of v and the fingers
in the direction of B. The palm will the face the direction of the force (and
hence the deflection) if the particle has a positive charge. The results are
(a)
tow ard top ofpage
(b)
outofthe page , since the charge is
(d)
into the page
negative.
(c)
zero force
Problem 4. A proton travels with a speed of 3.0 × 106 m/s at an angle of 37° with the
direction of a magnetic field of 0.30 T in the +y direction. What are (a) the magnitude of the
magnetic force on the proton and (b) the proton’s acceleration?
Solution: (a)
F = qvB sin θ
(
)
)(
= 1.60 × 10−19 C 3.0 × 106 m s ( 0.30 T ) sin ( 37° ) = 8.7 × 10−14 N
(b) a =
F 8.7 × 10−14 N
=
= 5.2 × 1013 m s2
m 1.67 × 10-27 kg
Problem 5. A proton moves perpendicularly to a uniform magnetic field B at 1.0 × 107
m/s and experiences an acceleration of 2.0 × 1013 m/s2 in the +x direction when its velocity is
in the +z direction. Determine the magnitude and direction of the field.
−27
13
2
F m a 1.67 × 10 kg 2.0 × 10 m s
Solution: B = =
=
= 0.021 T
qv qv
1.60 × 10−19 C 1.0 × 107 m s
(
(
)(
)(
)
)
The right hand rule shows that B must be in the -y direction to yield a force
in the +x direction when v is in the +z direction.
Problem 6. A current I = 15 A is directed along the positive x axis and perpendicularly to
a magnetic field. The conductor experiences a magnetic force per unit length of 0.12 N/m in
the negative y direction. Calculate the magnitude and direction of the magnetic field in the
region through which the current passes.
Solution: From F = BIL sinθ , the magnetic field is
B=
FL
0.12 N m
=
= 8.0 × 10−3 T
Isin θ (15 A ) sin 90°
The direction of B must be the + z direction to have F in the –y direction when I is
in the +x direction.
Problem 7. A wire carries a steady current of 2.40 A. A straight section of the wire is
0.750 m long and lies along the x axis within a uniform magnetic field of magnitude 1.60 T in
the positive z direction. If the current is in the + x direction, what is the magnetic force on the
section of wire?
Solution: The magnitude of the force is
F = BIL sin θ = (1.60 T ) ( 2.40 A ) ( 0.750 m ) sin ( 90.0°) = 2.88 N
and, from the right hand rule, the force is in the − y direction .
Problem 8. A wire carries a current of 10.0 A in a direction that makes an angle of 30.0°
with the direction of a magnetic field of strength 0.300 T. Find the magnetic force on a 5.00m length of the wire.
Solution: F = BIL sinθ = ( 0.300 T )(10.0 A )( 5.00 m ) sin ( 30.0°) = 7.50 N
Problem 9. A wire with a mass per unit length of 1.00 g/cm is placed on a horizontal
surface with a coefficient of friction of 0.200. The wire carries a current of 1.50 A eastward
and moves horizontally to the north. What are the magnitude and the direction of the smallest
vertical magnetic field that enables the wire to move in this fashion?
Solution: For minimum field, B should be perpendicular to the wire. If the force is
to be northward, the field must be directed dow nw ard .
To keep the wire moving, the magnitude of the magnetic force must equal
that of the kinetic friction force. Thus, BIL sin90° = μ k ( m g) , or
B=
μk ( m L) g ( 0.200) (1.00 g cm ) ( 9.80 m s2 ) ⎛ 1 kg ⎞ ⎛ 102 m ⎞
=
⎜⎝ 103 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T
Isin 90°
(1.50 A )(1.00)
Problem 10. A conductor suspended by two flexible wires as shown in the figure has a
mass per unit length of 0.040 0 kg/m. What current must exist in the conductor for the tension
in the supporting wires to be zero when the magnetic field is 3.60 T into the page? What is the
required direction for the current?
Solution: To have zero tension in the wires, the magnetic force per unit
length must be directed upward and equal to the weight per
unit length of the conductor. Thus,
Fm
L
= B I=
mg
, or
L
Bin
( m L) g = ( 0.040 kg m ) ( 9.80 m
I=
B
3.60 T
s2
)=
0.109 A
From the right hand rule, the current must be to the right if the force is to be
upward when the magnetic field is into the page.
Problem 11. A thin, horizontal copper rod is 1.00 m long and has a mass of 50.0 g. What
is the minimum current in the rod that can cause it to float in a horizontal magnetic field of
2.00 T?
Solution: If the rod is to float, the magnetic force must be directed upward and
have a magnitude equal to the weight of the rod. Thus, BIL sin θ = m g , or
I=
mg
B L sin θ
For minimum current, sin θ = 1 giving
m g ( 0.0500 kg) ( 9.80 m s )
=
= 0.245 A
Im in =
BL
( 2.00 T ) (1.00 m )
2
Problem 12. In the figure , the cube is 40.0 cm on each edge. Four straight segments of
wire—ab, bc, cd, and da—form a closed loop that carries a current I = 5.00 A, in the direction
shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction.
Determine the magnitude and direction of the magnetic force on each segment.
Solution:For each segment, the magnitude of the force is given by F = BIL sinθ
and the direction is given by the right hand rule. The results of applying these
to each of the four segments are summarized below.
Segment
L (m)
θ
F (N)
Direction
ab
0.400
180°
0
_
bc
0.400
90.0°
0.0400
negative x
cd
0.400 2
45.0°
0.0400
negative z
0.0566
parallel to x-z
plane at 45° to
both +x and +z
directions
da
0.400 2
90.0°
y
d
a
z
I
b
c
B
x
Problem 13. Find the direction of the current in the wire as shown in the figure that would
produce a magnetic field directed as shown, in each case.
Solution: Imagine grasping the conductor with the right hand so the fingers curl
around the conductor in the direction of the magnetic field. The thumb then
points along the conductor in the direction of the current. The results are
(a)
(b)
tow ard the left
outofpage
(c)
low er leftto upper right
Problem 14. At what distance from a long, straight wire carrying a current of 5.0 A is the
magnetic field due to the wire equal to the strength of Earth’s field, approximately 5.0 × 10–5
T?
Solution: From B = μ0I 2π r , the required distance is
μ I ( 4π × 10 T ⋅ m A ) ( 5.0 A )
= 2.0 × 10−2 m = 2.0 cm
r= 0 =
2π B
2π ( 5.0 × 10−5 T )
-7
Problem 15. The two wires shown in the figure carry currents of 5.00 A in opposite
directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic
field (a) at a point midway between the wires, (b) at point P1 (10.0 cm to the right of the wire
on the right), and (c) at point P2 (20.0 cm to the left of the wire on the left).
Solution: Assume that the wire on the right is wire 1 and that on the left is wire 2.
Also, choose the positive direction for the magnetic field to be out of the page
and negative into the page.
(a) At the point half way between the two wires,
⎡ μ0I1 μ0I2 ⎤
μ0
+
( I1 + I2 )
⎥=−
2π r
⎣ 2π r1 2π r2 ⎦
Bnet = − B1 − B2 = − ⎢
( 4π × 10 T ⋅ m A ) (10.0 A ) = − 4.00 × 10
2π ( 5.00 × 10 m )
−7
=−
−5
-2
T
or Bnet = 40.0 μT into the page
Bnet = + B1 − B2 =
(b) At point P1 ,
( 4π × 10
=
−7
Bnet
( 4π × 10
−7
Bnet =
)
T ⋅ m A ⎡ 5.00 A
5.00 A ⎤
−
= 5.00 μT outofpage
⎢
2π
⎣ 0.100 m 0.200 m ⎥⎦
Bnet = −B1 + B2 =
(c) At point P2 ,
μ0 ⎡ I1 I2 ⎤
⎢ − ⎥
2π ⎣ r1 r2 ⎦
)
μ0 ⎡ I1 I2 ⎤
⎢− + ⎥
2π ⎣ r1 r2 ⎦
T ⋅ m A ⎡ 5.00 A
5.00 A ⎤
⎢⎣ − 0.300 m + 0.200 m ⎥⎦
2π
= 1.67 μT outofpage
Problem 16. A wire carries a 7.00-A current along the x axis and another wire carries a
6.00-A current along the y axis as shown in the figure. What is the magnetic field at point P
located at x = 4.00 m, y = 3.00 m?
Solution: Call the wire along the x axis wire 1 and the other wire 2. Also, choose
the positive direction for the magnetic fields at point P to be out of the page.
At point P, Bnet = + B1 − B2 =
( 4π × 10
−7
Bnet =
or
μ0I1 μ0I2 μ0 ⎛ I1 I2 ⎞
−
=
−
,
2π r1 2π r2 2π ⎜⎝ r1 r2 ⎟⎠
)
T ⋅ m A ⎛ 7.00 A 6.00 A ⎞
−7
−
⎜⎝
⎟ = +1.67 × 10 T
2π
3.00 m 4.00 m ⎠
Bnet = 0.167 μT outofthe page
Problem 17. Two parallel wires are 10.0 cm apart, and each carries a current of 10.0 A.
(a) If the currents are in the same direction, find the force per unit length exerted by one of the
wires on the other. Are the wires attracted or repelled? (b) Repeat the problem with the
currents in opposite directions.
Solution:
( 4π × 10 T ⋅ m A ) (10.0 A )
F μ0II
1 2
=
=
2π ( 0.100 m )
L 2π d
−7
(a)
= 2.00 × 10−4 N m
2
( attraction)
(b) The magnitude remains the same as calculated in (a), but the wires are
repelled. Thus,
F
= 2.00 × 10−4 N m
L
( repulsion)
Problem 18. A wire with a weight per unit length of 0.080 N/m is suspended directly
above a second wire. The top wire carries a current of 30.0 A, and the bottom wire carries a
current of 60.0 A. Find the distance of separation between the wires so that the top wire will
be held in place by magnetic repulsion.
Solution: In order for the system to be in equilibrium, the repulsive magnetic force
per unit length on the top wire must equal the weight per unit length of this
wire.
Thus,
F μ0II
1 2
=
= 0.080 N m , and the distance between the wires will be
L 2π d
d=
μ0II
1 2
2π ( 0.080 N m
( 4π × 10
−7
)
=
)
T ⋅ m A ( 60.0 A )( 30.0 A )
2π ( 0.080 N m
)
= 4.5 × 10−3 m = 4.5 m m
Problem 19. In the following figure, the current in the long, straight wire is I1 = 5.00 A,
and the wire lies in the plane of the rectangular loop, which carries 10.0 A. The dimensions
are c = 0.100 m, a = 0.150 m, and A = 0.450 m. Find the magnitude and direction of the net
force exerted by the magnetic field due to the straight wire on the loop.
Solution: The magnetic forces exerted on the top and bottom segments of the
rectangular loop are equal in magnitude and opposite in direction. Thus,
these forces cancel and we only need consider the sum of the forces exerted
on the right and left sides of the loop. Choosing to the left (toward the long,
straight wire) as the positive direction, the sum of these two forces is
Fnet = +
μ0II
μ II A
μ II A ⎛ 1
1 ⎞
1 2A
− 012 = 012 ⎜ −
⎟,
2π c
2π ( c+ a)
2π ⎝ c c+ a⎠
( 4π × 10
=
−7
or Fnet
)
T ⋅ m A ( 5.00 A )(10.0 A )( 0.450 m ) ⎛
1
1 ⎞
−
⎜⎝
⎟
2π
0.100 m 0.250 m ⎠
= + 2.70 × 10−5 N = 2.70 × 10−5 N to the left
Problem 20. What current is required in the windings of a long solenoid that has 1 000
turns uniformly distributed over a length of 0.400 m in order to produce a magnetic field of
magnitude 1.00 × 10–4 T at the center of the solenoid?
Solution: The magnetic field inside a long solenoid is B = μ0nI= μ0 ⎛⎜⎝
N⎞
⎟ I. Thus,
L⎠
the required current is
(1.00 × 10 T) ( 0.400 m ) = 3.18 × 10−2 A = 31.8 m A
BL
=
μ0N ( 4π × 10-7 T ⋅ m A ) (1000)
−4
I=
Problem 21. It is desired to construct a solenoid that has a resistance of 5.00 Ω and that
produces a magnetic field at its center of 4.00 × 10–2 T when it carries a current of 4.00 A.
The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the
radius of the solenoid is to be 1.00 cm, determine (a) the number of turns of wire needed and
(b) the length the solenoid should have.
Solution: (a)
From R = ρL A , the required length of wire to be used is
−3
⎡
⎤
R ⋅ A ( 5.00 Ω) ⎣π ( 0.500 × 10 m ) 4⎦
=
= 57.7 m
L=
1.70 × 10−8 Ω ⋅ m
ρ
2
The total number of turns on the solenoid (i.e., the number of times this
length of wire will go around a 1.00 cm radius cylinder) is
N =
L
57.7 m
=
= 919
2π r 2π ( 1.00 × 10−2 m )
(b) From B = μ0nI, the number of turns per unit length on the solenoid is
n=
B
4.00 × 10−2 T
=
= 7.96 × 103 turns m
μ0I ( 4π × 10-7 T ⋅ m A ) ( 4.00 A )
Thus, the required length of the solenoid is
N
919 turns
=
= 0.115 m = 11.5 cm
n 7.96 × 103 turns m