PHY2010S: Electromagnetism for Engineers
Part II: Magnetism
WPS 1 solutions
Dr. James Keaveney
james.keaveney@uct.ac.za
Part 1 Magnetic Lorentz force on charged particles
Problem 1 A particle has a charge of 1.6 × 10−19 C and
initially has a velocity v of 1 × 108 ms−1 in the positive xdirection and no velocity in the y or z directions as shown in
the diagram. The particle travels through a region of space
with a uniform magnetic field of strength 0.1T pointing into
the page. The particle’s velocity vector makes an angle (θ)
of 90 degrees with the field lines. Assuming the particle loses
no energy and the magnetic field region extends to infinity
in all directions, select the option below which best describes
the movement of the particle in the field region.
Solution The magnetic Lorentz force law and the righthand-rule tell us that this (positively charged) particle will
initially experience a force pointing directly upwards. The
force vector will always point at 90 degrees to the particle’s
velocity vector, resulting in circular motion with constant
speed. In the diagram, an example trajectory is shown by
the red dots and dashed line.
Problem 2 What is the the magnitude of the magnetic
Lorentz force exerted on the particle in question 1?
1
Solution The magnitude is given by the Lorentz force law
|F~ | = q(v.B. sin θ) = (1.6×10−19 C)(1×108 ms−1 )(0.1T )(1) =
0.0000000000016N
Problem 3 What is the total work done on the charged
particle passing through the magnetic field by the magnetic
Lorentz force in question 1?
Solution The magnetic lorentz force never does any work
on charged particles as the force vector is always 90 degrees
~ =0
to the velocity vector, i.e, F~ · dl
Part 2 Electric currents
Problem 1 The section of cylindrical wire in the diagram
has length of 1m, cross sectional area (A) of 3.14mm2 and
a volume charge density (ρ) of 1.2 × 1010 C.m−3 . Electrons
flow up the wire with a velocity ~velectrons = 0.5mms−1 . The
wire section is surrounded by a uniform magnetic field of
magnitude 2T that points horizontally towards the negative
x direction and has no component in the y or z directions.
What is the magnitude of the total force on the wire section?
2
Solution 1 The Rtotal force on the volume current in the wire
~
is given by F~ = (~v × B)ρdτ
. This gives
Z
~
|F | =
(0.0005ms−1 · 2T · 1) · (1.2 × 1010 Cm−3 ) dτ
Z
=
(1.2 × 107 )dτ
integrating over the volume corresponds to multiplying by
the volume of the cylinder (π·r2 ·h = (3.14)·(0.000001)·(1)) =
3.14 × 10−6 . This gives
|F~ | = (1.2 × 107 ) · (3.14 × 10−6 ) = 37.6N
Solution 2 The volume current density J~ is equal to the
volume charge density ρ times the charge velocity ~v . This
gives J = (1.2 × 1010 C.m−3 ).(0.0005ms−1 ) = 6 × 106 Am−2 .
As J~ represents the current per unit area, we integrate J~ over
~
the cross
R sectional area of the wire to get the total current I
~
(I~ = S Jda).
Hence,
~ = (6 × 106 ) · (0.00000314) = 18.84A
|I|
The total force on a current is then given by
Z
~ × B)
~
~ = 18.84 · 1 · 2 = 37.6N
|F | = I (dl
Problem 2 In which direction does the current from the
previous question flow?
Solution Downwards. Current is defined as flowing in the opposite directions to the electrons.
Problem 3 A wire of length l = 2.5m carries a current of
magnitude 2A through a region where a uniform magnetic
field of 4T points directly upwards. The wire lies in the x − y
plane and experiences a force of magnitude 8.45N. What is
the smaller angle (θ) between the wire and the field lines?
3
Solution The total force on the wire is given by
Z
~ × B)
~
~ = I · l · B · sin θ
F = (dl
Therefore
8.45 = (2 · 2.5 · 4 · sin θ)
8.45
= sin θ
2 · 2.5 · 4
8.45
−1
θ = sin
2 · 2.5 · 4
θ = 25 degrees
Problem 4 In what direction does the total force vector
from the previous question point?
Solution The right-hand-rule tells us that the force vector
points directly out of the page.
Problem 5 A thin disc carries a uniform surface charge
density of 6 × 10−6 Cm−2 and spins with an angular velocity
ω = 10s−1 . The surface current density at a distance of 1cm
from the centre of the disc is ? Am−2 .
Solution The surface current density K is equal to the surface charge density σ times the velocity ~v . The linear velocity at r = 1cm is ω · r = 10 · 0.01 = 0.1ms−1 . Therefore
K = (6 × 10−1 ) · (0.1) = 0.0000001Am−1
4
Problem 6 A uniformly charged solid sphere, of
radius 0.01m and total charge 10−6 C, is centered
at the origin and spinning at a constant angular
velocity ω = 1s−1 about the z axis as shown in the
diagram. The volume charge density of the sphere
is ?? Cm−3 . The point p is located on the surface
of the sphere at the spherical coordinates (0.01,
45, 45) with the radial and angular coordinates
given in metres and degrees. The magnitude of
the current density J~ at the point p is ?? Am−2 .
Answer in standard decimal form (not scientific
notation).
Solution The volume charge density ρ is =
10−6
C
−3
= (4/3)·3.14·(0.01)
. The linear
3 = 0.238Cm
(4/3)πr3
velocity ~v at the point p is given by ~v = ωr sin θφ̂ =
(1) · (1 × 10−2 ) · (0.701)φ̂ = (7 × 10−3 )φ̂ ms−1 .
The current density is given by J = ρ · v =
(0.238) · (7 × 10−3 ) = 0.0016Am−2
5