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Centripetal force on charges in
magnetic fields
Which way does a particle get pushed if the the
magnetic field is is always perpendicular to the
direction of travel ?
• No mater which way the charged particle
turns the force on it is always perpendicular to
its motion.
Circular motion -the force is always
perpendicular to the direction of travel
-negatively charged particle
-magnetic field into the
-magnetic force is perpendicular to the velocity, and so velocity
changes in direction but not magnitude. Uniform circular motion
results.
electron moving at right angles to a
uniform magnetic field.
• Circular motion
• F = mv2 / r
• F  force (Newton, N)
• m mass ( Kg)
• v  velocity (m/s)
• B  magnetic field
• q charge
• Force on a charge particle
in a magnetic field
• F = qvB sinθ
• v perpendicular to B:
• F = qvB
• qvB = mv2 / r
• r = mv/ q B
• r = mv/ q B
• Gives you the radius of a charge particles
path in a magnetic field, given its mass and
velocity.
Relationship between radius and magnetic field,
mass and velocity
r = mv/ q B
• Magnetic field B: The stronger the magnetic field, the
stronger the force– and therefore the smaller the
radius of the charge
• Velocity v: the more speed a charged particles has, the
harder it is for the magnetic field to corral ( circle) the
particle, and so it travels in a circle with a bigger radius.
• Mass m: the more mass the charged particle has, the
harder it’ll be to bend its path, sot the more mass, the
bigger the radius of the circle travels in.
• If the velocity of the electron is due to its having been
accelerated through a potential difference of magnitude V
(volts), then the kinetic energy of the electron is
• ½ mv2 = qV
• write for the charge to mass ratio of the electron
q/m = 2V / B2r2
Example
• Alpha particles of charge q = +2e and mass m
= 6.6 x10-27 kg are emitted from a radioactive
source at a speed of 1.6 x 10 7 m/s. What
magnetic field strength would be required to
bend these these in a circular path of radius r
= .25 m?
• e = 1.6 × 10-19
Alpha particles of charge q = +2e and mass m = 6.6 x10-27 kg are emitted from a
radioactive source at a speed of 1.6 x 10 7 m/s. What magnetic field strength would
be required to bend these these in a circular path of radius r = .25 m?
e = 1.6 × 10-19
• Set the force on the particle due to the magnetic
field equal to centripetal force necessary to keep
the particle moving in a circle.
• qvb = mv2/r
• cancel the v's where possible
• qB = mv/r
• B = (mv)/(qr)
• = (6.6 x10-27 * 1.6 x107)/(2e * .25)
• = 1.32 T
Example
• A singly charged positive ion has a mass of
2.5 x 10-26 kg. After being accelerated
through a potential difference of 250 V, the
ion enters a magnetic field of 0.5 T, in a
direction perpendicular to the field. Calculate
the radius of the path of the ion in the field
A charged positive ion has a mass of 2.5 x 10-26 kg. After being accelerated
through a potential difference of 250 V, the ion enters a magnetic field of 0.5
T, in a direction perpendicular to the field. Calculate the radius of the path of
the ion in the field
q = 1.6x10 -19 C
m = 2.5x10
DV = 250V
B = 0.5T
-26
kg DV =
W DK
=
=
q
q
1
2mv
q
2
2DVq
2(250)(1.6x10 -19 ) 56,568 m/s
v=
=
=
-26
m
2.5x10
r =?
FB = Fc
We need to
solve for the
velocity!
mv 2
qvB =
r
mv
r=
qB
(2.5x10 -26 )(56, 568)
0.0177
m
r=
=
(1.6x10-19 )(0.5)