Magnetism

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Magnetism
Forces caused by Magnetic Fields
Magnetic Fields
•
•
•
•
•
Measured in Teslas (T) or Gauss (G)
10,000 G = 1 T
Moving charge causes magnetic fields
Remember the domain theory
If moving charge causes magnetic fields,
can a magnetic field affect a moving
charge?
Force on a Moving Charge
• A moving charge experiences a force in a
magnetic field
• F = qv x B
• The magnitude of the force is qvB sinθ
• The direction is determined by the Right
Hand Rule
Apply RHR
B
X
F
B
v
Determine the direction
of the force on the
charge if the charge is a
proton.
Determine the
direction of the
velocity of the
charge.
Apply RHR
v
v
B
X
F
B
F
Force on a Current Carrying Wire
• A current carrying wire experiences a force
•
•
•
•
in a magnetic field
dF = I dL x dB
F = IL x B
The magnitude of the force is IlB sinθ
The direction is determined by the RHR
Problem 1
• A wire carrying a current of 2.0 A lies along
the x-axis. The current flows in the positive
x direction. A magnetic field of 1.2 T is
parallel to the xy plane and makes an angle
of 30º with the x-axis (pointing into the 1st
quadrant.) What is the force on a segment of
wire 0.40 m long?
Solution
• F = IlB sinθ
• F = (2.0 A)(0.40 m)(1.2 T) sin 30º
• F = 0.48 N k
Problem 2
• A wire carrying current of 1.5 A lies in a
horizontal surface in the xy plane. One end
of the wire is at the origin and the other is at
(3m,4m). The wire follows an erratic path
from one end to the other. A magnetic field
of 0.15T directed vertically downward is
present. What is the magnetic force acting
on the wire?
Solution
•
•
•
•
•
F = ILB sin θ
F = 0.15T (1.5A) (3i-4j)
F = 0.225 √(32 – 42)
F = 1.13 N
This is the magnitude only!
Homework
• Problems Ch. 29
• #1-5 and #33-38
F = qv x B
• Magnetic force on a charge at rest is zero.
• The magnetic force on a charge moving parallel or
anti-parallel to the magnetic field is zero.
• Magnetic force, when non zero is perpendicular to
velocity and magnetic field. Thus it contributes
only to a radial acceleration and never performs
work on an object or changes its speed (it can only
change its direction).
Continued
• Magnetic force is proportional to the charge
of a point charge and thus exerts forces only
on charged objects.
• The direction can be obtained by the righthand rule (positive charge only—reverse the
direction of the velocity for a negative
charge).
Tesla
• T = N/(Cm/s)
• T = N/Am
Change of direction
•
•
•
•
•
•
•
F = qvB
This force causes circular motion
F = mv2/r
mv2/r = qvB
B = mv/qr
Or m/q = Br/v (mass to charge ratio)
Mass spectrometry allows us to find the mass to
charge ratio
Three types of Motion
• When a charged particle moves into a magnetic
field three types of motion can occur:
• Move parallel to B and experience no force
• Move perpendicular to B supplying a centripetal
force
• Move at some angle to B and have components in
the parallel and perpendicular direction producing
helical or corkscrew pattern of motion
Force on a Current Carrying Wire
•
•
•
•
•
F = qv x B
F = e(Nv0) x B
N = # charges
N = n(volume) = nlA
Where l is the length
of the wire and A is
the cross-sectional
area
• F = nlA evo x B
•
•
•
•
•
I = JA
I = nevDA so
F = Il x B
dF = i dl x B
F = ∫idl x B= i∫dl x B
Force on a Semicircle of Wire
• A wire bent into a semicircle of radius R
forms a closed circuit and carries a current
I. The wire lies in the xy plane and a
uniform magnetic field is directed along the
positive y axis. Find the magnitude and
direction of the magnetic force acting on the
straight portion of the wire and on the
curved portion.
Answer
• F1 acting on the
straight portion has a
magnitude of F1 =
IlBsinθ = I(2R)B sin
90°
• F1 = 2IRB
• Direction out of the
page or +z axis
B
Continued
• F2 acting on the
curved part—We must
first write an equation
for dF2 on the length
ds
• dF2 = I ds X B = IB
sin θ ds
• dF2 = IB sin θ ds
• s = R θ so ds = r d θ
B
Continued
•
•
•
•
•
θ = 0 to θ = π
dF2 = IRB sin θ d θ
F2 = IRB ∫ sin θ d θ
From 0 to π
F2 = IRB (-cos θ) from 0
to π
• = -IRB (cos π – cos 0)
• = -IRB (-1-1) = 2IRB
• Direction: into the page or
–z axis
B
Torque on a Current Carrying
Loop
I
• Torque = F x r
• Torque max = Frsinθ
• Torque max = F2
(b/2)sin 90 +
F4(b/2)sin 90 =
IaB(b/2) +IaB(b/2) =
(IAB)/2 + (IAB)/2 =
IAB
• Where A = ab
B
1
a
2
b
3
4
Continued
I
• Ends or part 1 and 3
• Torque = F1 (a/2)sinθ
+ F3(a/2)sinθ
• Torque = IbB(a/2)sinθ
+ IbB(a/2)sinθ =
IabBsinθ = IAB sinθ
B
1
a
2
b
3
4
Magnetic moment
•
•
•
•
Torque = I(A x B)
μ = IA
Torque = μ x B
And U = - μ dot B
• A = Area Vector
Problem
• A rectangular coil of dimensions 5.40 cm x
8.50 cm consists of 25 turns of wire and
carries a current of 15.0 mA. A 0.350T
magnetic field is applied parallel to the
plane of the loop. A) Calculate the
magnitude of its magnetic dipole moment.
B) What is the magnitude of the torque
acting on the loop?
Answer
• μ = NIA = 25 (15 x 10-3 A)(0.0540
m)(0.0850 m) = 1.72 x 10-3 Am2
• Torque = 1.72 x 10-3 Am2 (0.350T) =
6.02x10-4 Nm

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