Version 001 – HW#5 - Magnetism – arts – (00224)
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Charged Particle in a Field 02
001 10.0 points
A charged particle is projected with its initial
velocity parallel to a uniform magnetic field.
What is the resulting path?
1. straight line parallel to the field. correct
2. straight line perpendicular to the field.
3. circular arc.
4. spiral.
5. parabolic arc.
Explanation:
The force on a moving charge due to a
magnetic field is given by
~ = q~v × B
~.
F
~ are parallel, then
If ~v and B
~ = 0.
~v × B
Hence the force on the particle is zero, and
the particle continues to move in a straight
line parallel to the field.
Field Direction
002 10.0 points
The direction of the magnetic field in a certain
region of space is determined by firing a test
charge into the region with its velocity in
various directions in different trials.
The field direction is
1. the direction of the velocity when the
magnetic force is a maximum.
2. perpendicular to the velocity when the
magnetic force is zero.
3. None of these correct
4. the direction of the magnetic force.
1
5. one of the direction of the velocity when
the magnetic field is zero.
Explanation:
The magnetic field direction can be determined according to the path of a charge which
moves in the field. Since
~ = q ~v × B
~ ,
F
the component of velocity along the magnetic
field doesn’t contribute to the magnetic force;
i.e., a charge moving along the magnetic field
will keep its motion status. On the contrary,
if the initial velocity in perpendicular to the
magnetic force, there will be a centripetal
force so that the charge will undergo circular
motion.
Proton in a Magnetic Field 02
003 10.0 points
A proton moving at 6.9 × 106 m/s through a
magnetic field of 2.4 T experiences a magnetic
force of magnitude 1.2 × 10−12 N.
What is the angle between the proton’s
velocity and the field? The charge on a
proton is 1.60218 × 10−19 C and its mass is
1.67262 × 10−27 kg.
Correct answer: 26.8903◦ .
Explanation:
Let :
E
B
v
qp
= 1.2 × 10−12 N ,
= 2.4 T ,
= 6.9 × 106 m/s ,
= 1.60218 × 10−19 C ,
and
mp = 1.67262 × 10−27 kg .
The Lorentz force acting on a moving
charged particle in a magnetic field is
F = q v B sin θ
F
θ = arcsin
qvB
1.2 × 10−12 N
= arcsin
1.60218 × 10−19 C
Version 001 – HW#5 - Magnetism – arts – (00224)
1
×
(6.9 × 106 m/s) (2.4 T)
2
= 26.8903◦ .
i2
i1
Alpha Particle
004 10.0 points
An alpha particle has a mass of 6.6 × 10−27 kg
and is accelerated by a voltage of 0.592 kV.
The charge on a proton is 1.60218×10−19 C.
If a uniform magnetic field of 0.065 T is
maintained on the alpha particle and perpendicular to its velocity, what will be particle’s
radius of curvature?
Correct answer: 0.0759739 m.
Explanation:
Find the direction of the magnetic force on
i2 due to the magnetic field of i1 .
1. to the left
2. to the right correct
3. out of the paper
4. to the right and downward
Let : B = 0.065 T ,
V = 0.592 kV = 592 V ,
m = 6.6 × 10−27 kg , and
q = 2 e = 3.20435 × 10−19 C .
From Newton’s second law,
F = qvB =
5. to the left and upward
6. into the paper
7. to the left and downward
8. to the right and upward
m v2
r
qBr
.
v=
m
Explanation:
i1
i2
The kinetic energy is
K=
q2 B 2 r2
1
m v2 =
= qV ,
2
2m
so the particle’s radius of curvature is
s
1 2V m
r=
B
q
s
2 (592 V) (6.6 × 10−27 kg)
1
=
0.065 T
3.20435 × 10−19 C
= 0.0759739 m .
Magnetic Force Directions 02
005 10.0 points
Two parallel wires carry opposite current as
shown.
B1
F2
Wire 1, which carries a current i1 , sets up a
~ 1 , which points into the paper
magnetic field B
~1
at the position of wire 2. The direction of B
is perpendicular to the wire, as illustrated
below. The magnetic force on a length l
~ 2 = i2 ~l × B
~ 1 . Since i2 flows
of wire 2 is F
~ 1 is to the right.
downward, ~l × B
Parallel Wires 01
006 10.0 points
Two parallel wires carry equal currents in
Version 001 – HW#5 - Magnetism – arts – (00224)
the opposite directions. Point A is midway
between the wires, and B is an equal distance
on the other side of the wires.
A
B
3
the same distance from both wires, the other
wire gives a magnetic field at A of the same
magnitude, also directed into the paper due
to the right-hand rule. The total magnetic
field at A is
BA = 2 Bup,A =
2 µ0 I
.
πr
Now, the field at B due to the upgoing wire is
Bup,B =
What is the ratio of the magnitude of the
magnetic field at point A to that at point B?
BA
2
=
BB
3
B
4
2. A =
BB
3
B
3. A = 0
BB
B
4. A = 4
BB
B
5. A = 2
BB
B
6. A = 3 correct
BB
5
B
7. A =
BB
2
B
1
8. A =
BB
2
B
10
9. A =
BB
3
B
1
10. A =
BB
3
Explanation:
The magnetic field due to a long wire is
µ0 I
.
B=
2πr
Let the distance between the wires be r.
The magnetic field at A due to the upgoing
wire is
µ0 I
µ0 I
Bup,A =
=
.
2 π (r/2)
πr
1.
The right-hand rule tells us the direction is
into the paper. Due to the fact that A is
µ0 I
µ0 I
=
,
2 π (3 r/2)
3πr
again into the paper, while
Bdown,B =
µ0 I
µ0 I
=
2 π (r/2)
πr
out of the paper. So
BB = Bdown,B − Bup,B
1
µ0 I
1−
=
πr
3
2 µ0 I
,
=
3πr
out of the paper. Comparing magnitudes, we
find
2 µ0 I
BA
= πr = 3 .
2 µ0 I
BB
3πr
Long Solenoid
007 10.0 points
What current is required in the windings of a
long solenoid that has 1250 turns uniformly
distributed over a length of 0.578 m in order
to produce inside the solenoid a magnetic field
of magnitude 6.05×10−5 T? The permeablity
of free space is 1.25664 × 10−6 T m/A.
Correct answer: 22.262 mA.
Explanation:
Let : N = 1250 ,
ℓ = 0.578 m , and
B = 6.05 × 10−5 T .
Version 001 – HW#5 - Magnetism – arts – (00224)
Magnetic field inside the solenoid is
Basic Concepts:
Charged Particle:
µ0 N I
, then
ℓ
Bℓ
I=
µ0 N
(6.05 × 10−5 T) (0.578 m)
=
µ0 (1250)
B=
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
~ = q ~v × B.
~
Solution: The force is F
~
B = B −k̂ ,
keywords:
Particle Deflection 01
008 10.0 points
A negatively charged particle moving parallel
to the x-axis enters a magnetic field (pointing
into of the page), as shown in the figure below.
y
×
×
×
~ ×
×
B
×
×
−q
v
×
×
×
×
×
×
×
z
x
×
×
~
B
×
×
×
×
×
×
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
What is the initial direction of deflection?
~v = v (+ı̂) , and
q < 0 , therefore,
~ = −|q| ~v × B
~
F
h
i
= −|q| v B (+ı̂) × −k̂
= −|q| v B (+̂)
b = −̂ .
F
This is the third of eight versions of the
problem.
Particle Deflection 04
009 10.0 points
A positively charged particle moving parallel to the z-axis enters a magnetic field (pointing toward the bottom of the page), as shown.
ı̂ is in the x-direction, ̂ is in the y-direction,
and k̂ is in the z-direction.
x
~
B
b = −̂ correct
1. F
y
~ = 0 ; no deflection
2. F
v
b = −k̂
3. F
b = +̂
4. F
b = +ı̂
5. F
b = +k̂
6. F
b = −ı̂
7. F
Explanation:
Magnetic Force on a
~ = q ~v × B
~
F
= 22.262 mA .
×
4
z
+q
~
B
What is the initial direction of deflection?
b = +̂ correct
1. F
b = −k̂
2. F
b = +k̂
3. F
Version 001 – HW#5 - Magnetism – arts – (00224)
5
~ = 0 ; no deflection
4. F
~ = 0 ; no deflection correct
6. F
b = −̂
5. F
b = −̂
7. F
b = −ı̂
6. F
b = +ı̂
7. F
Explanation:
~
~
~
The force is F = q ~v × B , B = B −k̂ ,
1
~v = √ v −k̂ + ı̂ , and q > 0 , so
2
~ = +|q| ~v × B
~
F
h i
= +|q| v B −k̂ × (−ı̂)
= +|q| v B (+̂)
b = +̂ .
F
Particle Deflection 07
010 10.0 points
A negatively charged particle moving parallel
to the y-axis enters a magnetic field (pointing
toward the right-hand side of the page), as
shown. ı̂ is in the x-direction, ̂ is in the
y-direction, and k̂ is in the z-direction.
z
Explanation:
~ = q ~v × B
~, B
~ = B (+̂) ,
The force is F
1
~v = √ v +̂ + k̂ , and q < 0 , so
2
~ = −|q| ~v × B
~
F
= −|q| v B [(+̂) × (+̂)]
= −|q| v B (0)
b = 0 no deflection .
F
Particle Deflection 08
011 10.0 points
A negatively charged particle moving at 45◦
angles to both the z-axis and x-axis enters
a magnetic field (pointing towards the righthand side of the page), as shown in the figure.
ı̂ is in the x-direction, ̂ is in the y-direction,
and k̂ is in the z-direction.
x
~
B
y
z
~
B
x
−q
y
~
B
v
~
B
What is the initial direction of deflection?
b = −ı̂
1. F
b = +k̂
2. F
b = +̂
3. F
b = −k̂
4. F
b = +ı̂
5. F
v
−q
What is the initial direction of deflection?
b = √1
1. F
−k̂ + ı̂
2
1 b
+̂ − k̂
2. F = √
2
b = √1
3. F
−̂ + k̂
2
1
b=√
4. F
+k̂ − ı̂
2
1 b
5. F = √
+̂ + k̂
2
~ = 0 ; no deflection
6. F
Version 001 – HW#5 - Magnetism – arts – (00224)
b = +̂ correct
7. F
b = √1 (+̂ − ı̂)
8. F
2
b = −̂
9. F
b = √1 (−̂ + ı̂)
10. F
2
Explanation:
~ = q ~v × B
~ ,B
~ = B +k̂ ,
The force is F
1
~v = √ v −k̂ + ı̂ , and q < 0 , so
2
~ = −|q| ~v × B
~
F
h
i
1
= −|q| √ v B −k̂ + ı̂ × +k̂
2
1
= −|q| √ v B (+̂)
2
b = +̂ .
F
6