Magnetic field generated by
A long, straight current
A current loop
B=μoI/2πr
>
B=μoI/2r
A current generates magnetic field
I
r
1. Which one of the following statements concerning the effect of the
magnetic force on the wire is true?
(a) The wire will be pushed to the left.
(b) The wire will be pushed to the right.
>
>
<
<
×
B
A long straight vertical segment of wire traverses a magnetic field of
magnitude 2.0 T in the direction shown in the diagram. The length of the
wire that lies in the magnetic field is 0.060 m. When the switch is closed,
a current of 4.0 A flows through the wire from point P to point Q.
N
B
r
I
S
A loop of wire with a weight of 1.47 N is oriented vertically
and carries a current I = 1.75 A. A segment of the wire
passes through a magnetic field directed into the plane of
the page as shown. The net force on the wire is measured
using a balance and found to be zero. What is the
magnitude of the magnetic field?
(a) zero tesla
(d) 1.5 T
(b) 0.51 T
(e) 4.2 T X
(c) The wire will have no net force acting on it.
(d) The wire will be pushed downward, into the plane of the screen.X
(e) The wire will be pushed upward, out of the plane of the screen.
(c) 0.84 T
2. What is the magnitude of the magnetic force acting on the wire?
(a) 0.12 N
(c) 0.48 NX
(b) 0.24 N
(d) 67 N
(e) 0 N
Two loops carry equal currents I in the same direction. The loops are
held in the positions shown in the figure and are then released. Which
one of the following statements correctly describes the subsequent
behavior of the loops?
Two long, straight, parallel wires separated by a distance d carry currents
in opposite directions as shown in the figure. The bottom wire carries a
current of 6.0 A. Point C is at the midpoint between the wires and point O
is a distance 0.50d below the 6-A wire as suggested in the figure. The
total magnetic field at point O is zero tesla.
(a) Both loops move to the left.
(b) The loops remain in the positions shown.
(c) The top loop moves to the right; the bottom loop
moves to the right.
(d) The loops repel each other.
(e) The loops attract each other. X
1. Determine the value of the current, I, in the top
wire.
(a) 2 A
since
(c) 6 A
(e) This cannot be determined
(b) 3 A
(d) 18 A
specified.
X
the value of d is not
2. Determine the magnitude of the magnetic field at
point C if d = 0.10 m.
(a) 2.4x10–5 T
(b)
4.8x10–5
T
(c) 9.6x10–5 T
(d)
1.1x10–4
T
X (e) 1.4x10–4 T
Two long, straight wires are perpendicular to the plane of the paper as
shown in the drawing. Each wire carries a current of magnitude I. The
currents are directed out of the paper toward you. Which one of the
following expressions correctly gives the magnitude of the total
magnetic field at the origin of the x,y coordinate system?
(a) μoI/2d
(b) μoI/2πd
(c) μoI/√2πd
(d) μoI/√2d
(e) μoI/d
X
The radius of a coil of wire with N turns is r = 0.22 m. A current Icoil =
2.0 A flows clockwise in the coil, as shown. A long, straight wire
carrying a current Iwire = 31 A toward the left is located 0.05 m from
the edge of the coil. The magnetic field at the center of the coil is
zero tesla. Determine N, the number of turns.
(a) 2
(b) 4
X
(c) 3
(d) 1
(e) 6
“Loops” for Real Magnets
Magnetic field inside a solenoid
B=NμoI/l=nμoI
where N is number of loops or turns, l the length of the
solenoid, and n the number of the loops per unit length
Bitter Disk
Florida-Bitter Disk
B=nμoI
The larger I, the stronger B
Example: A solenoid 20 cm long and 4 cm in diameter is wounded with a
total of 200 turns of wire. The solenoid is aligned with its axis parallel
to the earth’s magnetic field which is 3x10-5 T in magnitude. What
should the current in the solenoid be in order for its field to exactly
cancel the earth’s field inside the solenoid?
The strongest field is
45 T, more than a
million times stronger
than the earth’s field,
3×10-5 T
B=nμoI
The larger n, the stronger B
Solution: Since l=0.2 m and N=200 turns, B=NμoI/l
I=Bl/μoN=(3x10-5T)(0.2)/(4πx10-7Tm/A)(200)=0.024 A=24 mA
The solenoid diameter has no significance except as a check that the
solenoid is long relative to its diameter.
Magnetic Force
F=BqvsinΘ for a moving charge
Applications of Magnetism
F=BIlsinΘ
@
&
I
×
Top view
•
b
B
F
I
F
&
Side view
r
a/2
a/2
Torque on a Current Loop
DC motor
F=IbB
τ=F×r
for a current
@
B
F
F
Θ
&
B
@
Side view
Loop plane || B
τ=Fa/2 + Fa/2 = Fa=IbBa=IBA
where A=ab, the area of the loop
Loop plane makes an angle Θ with B
τ=IBAsinΘ (because a/2sinΘ)
F
F
Loop plane makes an angle Θ with B
B τ=0
&
@
Side view
F
A current loop in a magnetic field always tends to turn so that its
plane becomes perpendicular to the field
Question: A current-carrying wire is in a uniform
magnetic field with the direction of the current
the same as that of the field
(a) There is a force on the wire that tends to move it
parallel to the field
(b) There is a force on the wire that tends to move it
perpendicular to the field
(c) There is a torque on the wire that tends to rotate
it until it is perpendicular to the field
(d) There is neither a force nor a torque on the wire
Answer: d
Question: A current-carrying loop in a magnetic
field always tends to rotate until the plane of the
loop is
(a) parallel to the field
(b) perpendicular to the field
(c) either parallel or perpendicular to the field,
depending on the direction of the current
(d) at a 45° angle with the field
Answer: b
Question: The nature of the force responsible
for the operation of an electric motor is
(a) electric
(b) magnetic
(c) a combination of electric and magnetic
(d) either electric or magnetic depending on
the design of the motor
Answer: b
Galvanometers
τ=INBAsinΘ=kφ (τ=kφ restoring toque)
φ= INBAsinΘ/k
The Hall Effect
××××××××××
+- + + + + + + +
EH
××××××××××
----------××××××××××
l
evdB =eEH so EH=vdB
where vd is the drift velocity
Hall emf (potential difference)
EH=Ehl= vdBl
Hall probe, density of charge carriers
Loudspeakers
As the current alternates at a frequency of the audio signal,
the speaker cone moves back and forth at the same frequency,
generating sound waves.
Mass Spectrometer
An electron travels through a region of space with no acceleration. Which
one of the following statements is the best conclusion?
(a) Both E and B must be zero in that region.
(b) E must be zero, but B might be non-zero in that region.
2r
X (c) E and B might both be non-zero, but they must be mutually
•••••
•••••
perpendicular.
(d) B must be zero, but E might be non-zero in that region.
(e) E and B might both be non-zero, but they must point in opposite
directions.
FE=qE, FM=qvB
Stage 1: qE=qvB thus v=E/B
Stage 2: qvB’=mv2/r so m=qB’r/v=qBB’r/E
Two charged particles are traveling in circular orbits with the same speed in
a region of uniform magnetic field that is directed into the page, as shown.
The magnitude of the charge on each particle is identical, but the signs of
the charges are unequal.
Which one of the entries in the table below is correct?
Mass Relationship
Sign of charge Q1
Sign of charge Q2
(a) m1 = m2
+
–
(b) m1 > m2
–
+
(c) m1 < m2
–
+
(d) m1 > m2
+
–
(e) m1 < m2
+
–
X
Magnetic Force
Moving charges in magnetic field: F=qvB
F=qvB
F=qvB is at the heart of all these applications!
Questions: Two ions have the same mass, but one is singly ionized and
other is doubly ionized. How will their positions on the film of the
mass spectrograph?
Answer: The two ions have the same v regardless of charge, and
F=mv2/R=qvB >> R=mv/qB
So the singly-ionized ion will end up at twice the distance of the doublyionized ion at the end of the half circle path.