 Physics
 Comb. Sci.
NSS Physics in Life In-class Worksheets
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27 Motion of Charged Particles in a Magnetic Field
27.1 Electric current and electron drift velocity
Learning objectives

Derive the relation I = nAvQ between electron drift velocity and electric current.
Key ideas

In the absence of electric field, the free electrons in an isolated metal wire move
positive ions
randomly and collide with the _______________________
in the wire.

In the presence of electric field, the electrons experience electric forces and drift
opposite
slowly in the _______________
direction of the electric field at the
drift
_______________
velocity.

The current carried by a conductor can be expressed as
I = nAvQ

drift
The _______________
velocity (~10−5 m s−1) of free electrons is extremely small
compared with their mean speed (~106 m s−1).

Electric field
_______________________
is set up along the circuit as soon as the circuit is
closed and all free electrons start to circulate around. Electrical energy is then
converted into other forms of energy in the loads, regardless of the magnitude of
the drift velocity.
Follow-up
Checkpoint ( p.319)
Exercise ( p.320)
ISBN: 9789880059414
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Example 27.1A
 Physics
 Comb. Sci.
Electric current ( p.319)
Iron has 1.7  1029 electrons per cubic metre. Estimate the size of current produced if
electrons in an iron wire of diameter 1 mm drift at about 104 m s1. The magnitude of
the charge of an electron is 1.60 × 10−19 C.
Solution
Applying I = nAvQ,
_____________________________________________________________________
I = 1.7  1029    (10−3/2)2  10−4  1.6 10−19 ≈ 2.14 A
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
2.
ISBN: 9789880059414
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 Physics
 Comb. Sci.
NSS Physics in Life In-class Worksheets
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27.2 Magnetic force on a moving charge
Learning objectives

Represent the magnetic force on a moving charge by F = BQv sin .
Key ideas

The magnitude of the magnetic force on a moving charged particle in a magnetic
field is given by
F = BQv sin 

The direction of the magnetic force on the charged particle can be determined by
Fleming’s left hand rule
____________________________________.

To pass through the crossed fields in a velocity selector without deflection, the
speed of the particles must be
v 

E
B
The motion of a charged particle in a uniform magnetic field depends on the angle
between its initial velocity and the direction of the field, which can be
summarized as

values of 
type of motion
= 0° or 180°
rectilinear motion
______________________________
= 90°
circular motion
______________________________
0° << 90° or 90° << 180°
helical motion
______________________________
In a mass spectrometer, the radii of the semi-circular paths taken by the charged
charge to mass
particles depend on their ______________________________
ratios, so that
different particles can be separated and identified.
ISBN: 9789880059414
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NSS Physics in Life In-class Worksheets
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 Physics
 Comb. Sci.
Follow-up
Checkpoint ( p.326)
Checkpoint ( p.330)
Exercise ( p.330)
4.
ISBN: 9789880059414
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 Physics
 Comb. Sci.
Example 27.2A
NSS Physics in Life In-class Worksheets
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Magnetic force acting on a helium nucleus
( p.323)
A helium nucleus of charge +2e enters a uniform magnetic field at 1.5  107 m s−1. The
magnetic field has a magnitude of 0.8 T and points into the paper. The magnitude of
the charge of an electron is 1.60 × 10−19 C.
+2e
1.5  107 m s−1
(a) Find the magnitude and direction of the force acting on the nucleus initially.
(b) Sketch the path of the helium nucleus.
Solution
Applying F = BQv,
(a) _________________________________________________________________
magnitude of force acting on the helium nucleus = 0.8  2  1.6  10−19  1.5  107
_________________________________________________________________
= 3.84  10−12 N
_________________________________________________________________
The magnetic force acting on the helium nucleus points upwards.
_________________________________________________________________
_________________________________________________________________
(b)
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Example 27.3A
 Physics
 Comb. Sci.
Velocity selector ( p.323)
In the following velocity selector, the metal plates are 2 cm apart. The electric field
between the metal plates is controlled by the voltage across the plates. A pair of slab
magnets is used to provide a uniform magnetic field of magnitude 0.8 T. When the
voltage is adjusted to 40 V, an ion projected at right angle to the magnetic field passes
through the gap without deflection. Find the velocity v of the ion.
uniform
magnetic field
v
2 cm
Solution
electric field in the gap = V / d = 40 / 0.02 = 2000 V m
_____________________________________________________________________
−1
Since the ion does not deflect, the magnetic force balances the electric force. Thus,
_____________________________________________________________________
Bqv = qE
_____________________________________________________________________
v = E / B = 2000 / 0.8 = 2500 m s−1
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
6.
ISBN: 9789880059414
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 Physics
 Comb. Sci.
Example 27.4A
NSS Physics in Life In-class Worksheets
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Mass spectrometer ( p.328)
In a mass spectrometer, charged particles with the same velocity, 1.52  107 m s1, are
projected at right angles into a uniform magnetic field of magnitude 1.5 T.
(a) A particle of charge 3.2  1019 C moves in a path of radius 0.21 m in the
magnetic field. What is the mass of the particle?
(b) If another charged particle moves in the magnetic field with the same deflection,
what conclusion can be drawn?
Solution
magnetic force provides the centripetal acceleration for the circular motion.
(a) The
_________________________________________________________________
mv2/r = BQv
_________________________________________________________________
m = rQB/v
_________________________________________________________________
= 0.21  3.2  10−19  1.5 / 1.52  107 ≈ 6.63  10−27 kg
_________________________________________________________________
_________________________________________________________________
(b) Since r = mv/QB and particles entering a mass spectrometer have the same velocity, particles
with
the same deflection have the same charge to mass (Q/m) ratio.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
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27.3
 Physics
 Comb. Sci.
Hall effect
Learning objectives

Know about Hall effect and derive the equation VH 

Examine magnetic fields using a Hall probe.
BI
for Hall voltage.
nQb
Key ideas

When a current passes through a conductor placed in a uniform magnetic field,
magnetic force
each of the charge carriers experiences a ______________________
and deflects
to the surfaces. The deflection of the moving charged carriers leads to an excess
or a deficiency of ______________________
on the upper or lower surface of the
charge carriers
conductor.

A p.d. is developed across the conductor due to the deflected charge carriers.
Each charge carrier moving in the conductor experiences an electric force that
opposes
balance
_______________
the magnetic force on it. These two forces ________________
each other in the steady state.
8.
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 Physics
 Comb. Sci.

NSS Physics in Life In-class Worksheets
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Hall effect
The ___________________
is the production of a Hall voltage across the
opposite surfaces of a current-carrying conductor placed in a magnetic field,
which is given by
VH 

BI
nQb
Hall probe
A ___________________
is a device based on the Hall effect to measure a
magnetic field.
Follow-up
Checkpoint ( p.338)
Checkpoint ( p.340)
Exercise ( p.340)
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Example 27.5A
 Physics
 Comb. Sci.
Hall probe ( p.336)
The following shows a thin copper film inside a Hall probe. When a current of 2 A
flows through the film, the Hall voltage developed across the width of the film is
1.97 μV. The thickness and width of the copper film are 0.3 mm and 3 mm respectively.
The electron density of copper is 8.47  1028 m−3.
0.3 mm
2A
1.97μV
3 mm
B
(a) What is the magnitude of the magnetic field B?
(b) Estimate the average drift velocity of the electrons in the copper film.
Solution
Applying VH = BI/nQb,
(a) _________________________________________________________________
B = VH nQb/I
_________________________________________________________________
= 1.97  10−6  8.47  1028  1.6  10−19  0.3  10−3 / 2
_________________________________________________________________
≈ 4.00 T
_________________________________________________________________
_________________________________________________________________
Applying I = nAvQ,
(b) _________________________________________________________________
I = n(db)vQ
_________________________________________________________________
v = I /ndbQ
_________________________________________________________________
= 2 / (8.47  10  3  10  0.3  10  1.6  10 ) ≈ 1.64  10 m s
_________________________________________________________________
28
−3
−3
−19
−4
−1
_________________________________________________________________
10.
ISBN: 9789880059414
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