Right Hand Rule
Magnetic field is defined in terms of
the force on a moving charge
B=F/qvsinΘ for a moving charge
or F=qvxB
B=F/IlsinΘ
or F=lIxB
for a current
• Hold your right hand open
• Place your fingers in the
direction of B
• Place your thumb in the
direction of v
• The direction of the force
on a positive charge is
directed out of your palm
– If the charge is negative, the
force is opposite that
determined by the right
hand rule
Question: The right-hand rule for the direction
of the force on a charged particle in a
magnetic field applies:
(a) only to positive charges
(b) only to negative charges
(c) to both positive and negative charges
(d) only when the particle is moving parallel
to field.
Question: A charge particle moves
perpendicularly through a magnetic field.
The effect of the field is to change the
particle’s
(a) charge
(b) mass
(c) velocity
(d) energy
Answer: c
Answer: a
Two charged particles of equal mass are traveling in circular orbits in a
region of uniform, constant magnetic field as shown. The particles are
observed to move in circular paths of radii R1 and R2 with speeds v1 and
v2, respectively.
Magnetic Field of A Current
Force between Currents
As the figure shows, the path of particle 2 has a smaller radius than that
of particle 1. Which one of the following statements about this system is
false?
(a) |v1/Q1| < |v2/Q2|
X
(b) Particle 2 carries a positive charge.
(c) Particle 1 carries a negative charge.
(d) Neither particle gains energy from the magnetic field.
(e) The particle velocities have no components parallel to the magnetic
field.
Solenoid
Field around long, straight current
B~I/r
B=μoI/2πr (r<<length of the wire)
where μo=4πx10-7 T·m/A is
magnetic permeability of free space
I
r
×
Question: The magnetic field a distance d
from a long, straight wire is proportional
to
(a) d
(b) d2
(c) 1/d2
(d) 1/d
The direction of B is determined by the right-hand rule
Answer: d
Question: Charged particles shown move in the vicinity of a
current carrying wire. For each charged particle, indicate
the direction of magnetic force due to the magnetic field
generated by the wire (Arrow indicates the direction of
motion of the particle, + or – indicates the sign of the
charge.
Example: Find the magnetic field in air 1 cm from a wire that
carries a current of 1 A.
Solution: Since r=10-2 m, we have
B=μoI/2πr=(4πx10-7 T m/A)(1A)/2π10-2m=2x10-5 T
This is only a little smaller than the magnitude
of the earth’s magnetic field. For this reason
great care is taken aboard ships to keep
current-carry wires from magnetic compasses.
I
r
F
×
B
F
+
F
(e) out of the plane of the page X
(b) 1.9x 10–5 T
(d) 7.9x10–5 T
(e) 9.4x10–5 T
+Q1 F
12
I
L
F12 F21
2. What is the magnitude of the magnetic field at the point P?
(c) 2.5x10–5 T X
F
L
(c) toward the bottom of the page
(a) 1.3x10–5 T
I
+
(b) to the left of the page
×
B
-
r12
1. What is the direction of the magnetic field at point P due to the
current in the wire?
(d) into the plane of the page
B
F21 +Q2
A long, straight wire is carrying a current of 5.0 A in the direction
shown in the figure. The point P is 0.040 m from the wire.
(a) to the right of page
·
F12 F21
r
I
r
I
I
B1=μoI1/2πr B2=μoI2/2πr
The fields are perpendicular to the wire, i.e., sinΘ=1, Θ=90o
F21=I1LB2= μoI1I2L/2π
F12=I2LB1= μoI1I2L/2πr =F
L/2πr
F/L= μoI1I2/2π
/2πr
QUICK QUIZ 19.5
Definition of Ampere: Definition of Ampere:
If I1=I2=1A and r=1 m
If I1 = 2 A and I2 = 6 A in the figure below, which of the
following is true:
(a) F1 = 3F2, (b) F1 = F2, or (c) F1 = F2/3?
then F/L=2×10 10-7 N/m
1 C=1 A·s (Q=I·t)
Andre Ampere: French scientist (1775-1836)
QUICK QUIZ 19.5 ANSWER
(b). The two forces are an actionreaction pair. They act on different
wires, and have equal magnitudes but
opposite directions.
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.
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.
Magnetic field at the center of a current loop
B=μoI/2πr
B=μoI/2r
(a) 2.4x10–5 T
(c) 9.6x10–5 T
(b) 4.8x10–5 T
(d) 1.1x10–4 T
X (e) 1.4x10–4 T
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?
(a) Both loops move to the left.
>
(b) The loops remain in the positions shown.
I
(c) The top loop moves to the right; the bottom loop
B
r
×
B
>
>
<
<
r
I
moves to the right.
(d) The loops repel each other.
(e) The loops attract each other. X
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?
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.
B=nμoI
The larger n, the stronger B
Question: When a ferromagnetic substance is
inserted in a current-carrying solenoid, the
magnetic field is
(a) slightly decreased
(b) greatly decreased
(c) slightly increased
(d) greatly increased
Answer: d