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4.5) Magnetic Forces
1) For a current carrying wire you use F = L I x B. For a free charge you use F = Q v x B. Use the
An electron is moving through a uniform magnetic field that is down. What will happen to it if it is
moving: west?
up?
up and west?
In the first case it will feel a force to the north and will turn that way. The velocity stays perpendicular to
the force so it goes in a circle (clockwise looking down.) In the second case there will be no force
because the field and velocity are aligned. It will continue moving up. In the third case, it will turn and
move up in a helix.
2) Draw the forces and path of an electron moving to the right in the following magnetic fields.
Field into the page.
Field to the right.
Field Down
3) A magnet is brought near a black and white TV screen. Describe and explain what happens.
Inside, electrons are moving rapidly toward the screen. The magnetic field changes their directions.
4) Charged particles can move really fast. A velocity selector can be used to measure this.
a) There are two charged plates producing a uniform electric field that will exert a force up on a
positive particle moving to the right. There will be two circular coils producing a uniform magnetic
field that will exert a force down on the particle. Place the field lines on the sides of the box and the
current lines around the box. Make a 3-D model of this.
b) Draw a ¾ view and a side view below. Use coloured arrows to clearly show the current, magnetic
field, magnetic force, electric field and electric force. Assume a positive charge and negative charges
are on top. Red lines are the electric field lines and the electric force is up. This means that the
magnetic force must be down, so they can cancel. That means the magnetic field lines must point out
of the page. This means that the current must be counter clockwise.
c) The strengths of the fields are adjusted so that the magnetic force is balanced by the electric force.
Solve for v. Calculate the speed if the electric field is 11 kV/m and the magnetic field is 0.20 T.
qE = qvB
therefore, v = E/B = 11,000/0.20 = 55 km/s
5) The charge to mass ratio of an electron was measured by J.J. Thompson using magnetic and electric
fields. Watch http://www.youtube.com/watch?feature=endscreen&v=JB6YT7mm9JQ&NR=1
a) Draw a diagram of the apparatus and label the given values.
5,000 V across the horizontal parallel plates (separated by 5.6 cm), makes beam curve
5,000 V across the vertical parallel plates of the electron gun on the left.
(6 V is used to heat the wire to release electrons, not needed in calculation)
12 V is used to produce a current of (0.35 A) in the coil to produce a magnetic field.
Path of electron is black dashed line
Electric field is red
Magnetic field is green and out of page.
b) The magnetic field causes the electrons to turn when the electric field is off. Use F = ma for circular
motion and the magnetic force equation to form an equation for q/m.
qvB = m v2/r
q/m = v/Br
c) The speed can be measured using the electric and magnetic fields as shown in the question above.
Substitute the equation for the speed into the equation for q/m.
v = E/B (E is the electric field – not energy!)
q/m = E/(B2r)
d) How do you calculate the strength of the electric field? Substitute this into the formula for v.
E = V/d
v = V/Bd
e) How do you calculate the strength of the magnetic field? Substitute this into the formula for v.
B = I N/L
v = V/(dI N/L)
f) The video gives you all the necessary data to calculate q/m except the turns per length of the
electromagnet. Calculate what N/L must be using the universal constants for e, me and othe
permeability of free space (4 x 10-7 Tm/A).
N/L = V/(dI v)
V = 5, 000 V, d = 0.056 m,  = 4 x 10-7 Tm/A, I = 0.35 A
But what is v? Use conservation of energy at the electron gun. qV = ½ m v2
v = 4.19 x 107 m/s
Therefore N/L = 48 turns per cm.
6) Mass spectrometers use electric and magnetic fields to measure the masses of individual atoms and
molecules. Ions with a charge of -1 are accelerated by a potential difference of 250 V into a uniform
magnetic field of 1.00 T where their path curves with a radius of 1.35 cm.
a) Compare the set up to what Thompson used.
It is basically the same. However, we know the charge.
a) Use conservation of energy in the electric field to find an equation for mass.
½ m v2 = q V
v = sqrt (2 qV/m)
or v2 = 2qV/m
b) Use F = ma to find another equation for mass.
qvB = mv2/r
m = qBr/v
c) You have two equations with two unknowns – mass and speed. Solve for m.
m = qBr/v therefore m2 = q2 B2 r2 /v2 = q2 B2 r2/(2qV/m) = m q B2 r2/2V therefore m = q B2 r2/2V
d) Use the data above to calculate the mass of the particle. (p. 398)
Textbook: 8.2, 8.3
p. 396 # 2-6, p.402 # 1 - 4, p. 435 # 10, 11, 16, 17