2. Today we are going to learn
• The formula for the force on a particle in a
magnetic field F = B Q v
• Circular path of particles; application in
devices such as the cyclotron and Cern
3. The force on a charged particle in a
magnetic field
F = B Q v
F = Force (N)
B = Magnetic field (T)
Q = charge (C)
V = velocity (m/s)
when the field is perpendicular to velocity.
4. Rules for a charged particle in a
magnetic field
No work is done by the magnetic field on the
particle. The particle only changes direction it
does not change its speed.
The charged particle moves in a circular path
with the force acting towards the centre of
curvature.
If the magnetic field is uniform the particle will
travel in a complete circle
5. Applying a magnetic field
circle if v
perpendicular
to B
spiral in a
dissipative
medium
helix if component of
v parallel to B
F
v
B
6. What will be the radius of the path
Force towards the centre F = BQv
Centripetal acceleration a = v2/r
Newton’s second law F = ma
BQv = mv2 /r
r = mv/BQ
http://www.vjc.moe.edu.sg/fasttrack/physics/VelocitySelector.htm
r
7. F = BIL, F = BQV
Q1A wire of length 40cm is placed in a field of 0.15T. The current in
the wire is 2.3A.
Calculate the force exerted on the wire.
Q2 An electron moving at 200km/s passes at 90o through a
magnetic field of flux density 0.025T. The electron has a charge of -
1.6x10-19C and a mass of 9.1x10-31-kg.
What force is exerted on the electron by the field?
What is the radius of the electron’s circular path? (hint: think ‘circular
motion’)
Q3 A wire of length 400mm is placed in a field of 1.5x10-1T. The
current in the wire is 2300mA.
Calculate the force exerted on the wire.
8. Q1 A wire of length 40cm is placed in a field of 0.15T. The current in the wire is 2.3A.
Calculate the force exerted on the wire.
F = Bil (1 mark)
F = 0.15 x 2.3 x 0.4 (1 mark)
F = 0.14N (1 mark)
Q2 An electron moving at 200km/s passes at 90o through a magnetic field of flux density 0.025T. The
electron has a charge of -1.6x10-19C and a mass of 9.1x10-31-kg. What force is exerted on the electron by
the field?
F = Bqv (1 mark)
F = 0.025 x 1.6x10-19x 200000 (1 mark)
F = 8x10-16N (1 mark)
What is the radius of the electron’s circular path? (hint: think ‘circular motion’)
r = mv2/F (1 mark)
r = 9.1x10-31 x (200000)2/ 8 x10-16 or Answer to part a (1 mark)
r = 4.9 x 10-5m (1 Mark)
Q3 A wire of length 400mm is placed in a field of 1.5x10-1T. The current in the wire is 2300mA.
Calculate the force exerted on the wire.
F = Bil (1 mark)
F = 0.15 x 2.3 x 0.4 (1 mark)
F = 0.14N (1 mark)
9. Cyclotron
http://www.youtube.com/watch?v=M_jIcDOkTAY
The cyclotron is used in hospitals to
produce high energy beams of particles.
It consists of 2 D shaped electrodes called
the ‘Dees’. The magnetic field is
perpendicular and an ac current is applied
to the Dees. The AC oscillator creates a
potential difference to accelerate the
particle. As the particle’s velocity increases
its radius will increase until it ready to hit a
target.
r = mv/BQ
10. The Mass Spectrometer
Mass spectrometers can be used
to separate charged particles.
Particles are fired through a slit,
and then into a magnetic field.
The lightest particles will bend
the most, the heaviest will bend
the least.
The particles can only get
through the slit is they are at the
correct velocity as there is a
collimator plate creating an
electric field before the slit and a
magnetic field creating a force on
the particle in opposite directions
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