Transcript Torque, Current Carrying Loop

Comprehension Check
1. An electron is moving at right angles to
uniform magnetic field; if the electron is
moving at .010c, determine the
magnitude of the force if:
1.
2.
3.
4.
B = 51 000nT (~Earth’s magnetic field)
B = 5.0mT (~fridge magnet)
B = 1.0T (~rare earth magnet)
B = 3.0T (~superconducting magnet)
Comprehension Check
1. F = 2.5x10-17N
2. F = 2.4x10-15N
3. F = 4.8x10-13N
4. F = 1.4x10-12N
Charged Particle Applications
and Torque
Physics 12
Force on a Current Carrying Wire
in a Magnetic Field
Take the equation for the force
experienced by a charged particle moving
in a magnetic field and the definition of
current to develop an equation for a
current carrying wire in a magnetic field
F  IB
Torque
Torque is the cross
product of radius and
force
The units are Nm but
are not joules!
The direction is
positive for ccw and
negative for cw
 
  r F

  rF sin 
Torque on a Current Carrying Loop
Galvanometer
The force experience by
a current carrying wire in
a magnetic field can be
used to build a
galvanometer
Based on the length of
wire, the strength of the
field, the tension in the
spring and the current,
different readings are
obtained
Charged Particle placed in a BField
When a charged particle is placed in a
magnetic field, it experiences a force
based on the cross product of its velocity
and the magnetic field intensity
Therefore, a charged particle experiences
no force if it is not moving
Circular Motion
When a charged particle is moving in a
magnetic field, it always experiences a
force that is at right angles to the velocity
This results in a change in the direction of
the velocity but not its magnitude
As a result, this force will provide a
centripetal acceleration towards the centre
of the circular path
How can we calculate centripetal
acceleration?
 r v

r
v
d  vt
r  vt
vt v

r
v
v 2 v

r
t
2
v
ac 
r
Centripetal Force
Like the centripetal
acceleration, the
centripetal force is always
directed towards the
centre of the circle
The centripetal force can
be calculated using
Newton’s Second Law of
Motion


F  ma
2
v
ac 
r
2
mv
Fc 
r
Charged Particle moving in an BField
A charged particle,
moving with an initial
velocity enters a
magnetic field as shown
in the diagram at the
right and will follow a
circular path as a result
of the b-field
We can solve this
problem through the use
the centripetal motion
Lorentz Force and Velocity
Selector
The Lorentz Force
gives the method of
calculating the total
force acting on the
charged particle
When the velocity is
equal to the ratio of
E/B, the particle
passes straight
through the fields,
otherwise it is
deflected

  
F  qE  qv  B
Mass Spectrometer
A mass
can be built using a
rB0
m spectrometer
 selector followed by an area of
velocity
q
v
uniform magnetic field. Particles that have
the correct velocity will pass through the
velocity selector and into the magnetic field
where they will be curved.
Develop an equation that will allow you solve
for the ratio m/q based on the magnetic field,
speed of the particle and radius of curvature
e/m
Thompson used a similar technique to
measure the ratio of charge to mass ratio
of electrons
Using variable electric and magnetic fields,
it is possible to investigate the charge of
the electron in comparison to its mass
Cyclotron
In a cyclotron, a charged
particle is accelerated in
an alternating electric field
and passes into an area of
uniform magnetic field
which turns it in a circle
and it passes back into the
magnetic field
As velocity increases, the
radius gets larger and
larger