DR / MOSTA

Chapter Nine
Electromagnetism
Magnetic fields
The magnetic field of the magnet is the region around a magnet where
magnetic effects can be experienced.
• The magnetic field is not visible but they
may be represented by magnetic field lines.
(or magnetic flux)
• The magnitude and the direction of the
magnetic field can be represented by its
magnetic flux density (B)
• The direction of the magnetic flux density at a point is that of the
tangent to the field line at the point
• The magnitude of the flux density is proportional to the number of
field lines per unit area.
• Although the magnetic field has been drawn it two dimensions, the
actual magnetic field is three dimensional.
• The magnetic field lines start at a north pole and end at a south pole
• The magnetic field lines are smooth curves which never touch or cross
• The strength of the magnetic field is indicated
by the distance between the lines-closer lines
means a stronger field.
If you have two bar magnets the magnetic field
appears as shown From the opposite figure we
can notice that The magnetic field due to one
magnet opposes that due to the other.
The magnetic field lines cannot cross and
consequently there is a point X known as
(Neutral point), where there is no resultant
magnetic field because the two fields are equal in
magnitude but opposite in direction.
1 PHYSICS
EDEXCEL
 DR / MOSTA

Magnetic fields and electric current

In 1890 Orested discovered that “a wire carrying an electric current
has an associated magnetic field” The size and shape of the magnetic
field depends on the size of the current and the shape of the conductor
through which the current is travelling Using the solenoid illustrated
in the next figure

The magnetic field due to a straight wire appears as a series of

concentric circles centered on the wire and the direction of the field
can be found by using the right –hand grip rule
Grasp the wire using the right hand with the thumb pointing in the
direction of the current, the figures then point in the direction of the
field

Anticlockwise Clockwise
2 PHYSICS
EDEXCEL
 DR / MOSTA

The force on a conductor in a magnetic field

A current –carrying wire is surrounded by a magnetic field. This
magnetic field interacts with an external magnetic field, giving rise to
a force on the conductor
In the opposite figure
The magnets create a uniform field, as soon as the
current in the copper rod is switched on; the rod
starts to roll, showing that a force is acting on it.
An experiment shows that the force is always
perpendicular to the plane, which contains
both the current and the external field at the
site of the conductor.
The direction o the force can be found by using
Fleming’s left rule “if the first and second
fingers and the thumb of the left hand are
placed comfortably, at right angles to each
other, with the first finger, pointing in the
direction of the field, and the second finger
pointing in the direction of the current, then the
thumb points in the direction of the force (the
direction which motion take place)”
We can explain this force by thinking about the magnetic fields of the
magnets and the current carrying the conductor; these fields combine

3 PHYSICS
EDEXCEL
 DR / MOSTA

and interact to produce the force on the rod The production of this
force is known as [the motor effect]
The strength of the magnetic field is known as the magnetic flux
density (you can imagine this quantity by represent the number of
magnetic field lines passing through a region per unit area) The
magnetic flux density is greater close to the bars of the magnets and
gets smaller as you move away from it
It defined as
F
B=
I×L
Angle bet. The
B is the magnetic flux density [measured in tesla] wine current &
the mag field
F is the force on the conductor [measured in Newton]
I is the current in the conductor [measured in Ampere]
L is the length of the conductor [measured in metre]
A magnetic flux density of 1 T produces a force of 1N on each meter
of wire carrying a current 1A at 90o to the field
Example (1)
The two wires beside are in a uniform magnetic field of 0.25T. Both
wires are 0.50m long and carry a current of 4.0 A, calculate the size
of the force on each wire.
(a) The wire is at 90o to the field
(b) The wire is at 30o to the field
Solution
B = 0.25T L = 0.50m I = 4.0 A F = ??
(a) The wire is at 90o to the field
F
B=
I×L
∴ F=I × L × B

= 4.0 x 0.50 x 0.25 = 0.50 N

4 PHYSICS
EDEXCEL
 DR / MOSTA

(b) The wire is at 30° to the field

F
B sinθ=
I×L
∴ F=I × L × Bsin θ

= 4.0 x 0.50 x 0.25 x sin 30° = 0.25 N

The force on a charged particle in a magnetic field
The magnitude of the force
F=B× Q× V ×sin θ

Where
F = the force on the particle (N)
B = the magnitude of the magnetic flux density of the field (T)
Q = the charge on the particle (C)
V = the magnitude of the velocity of the particle (ms-1)
From the last equation one can notice that a magnetic field cannot
exert a force on a stationary charged particle.
One tesla is the uniform magnetic flux density, which, acting
normally to a long straight wire carrying a current of 1 Ampere,
cause a force per unit length of 1 Nm-1 on the conductor.
* The direction on the force
The force on positively charged particle is the same as that on a
conductor, which is carrying a current in the same direction as that, in
which the particle is moving. It follows that the direction of the force
can be found by using Fleming’s left-rule.
Thus, if the particle shown in fig is positively charged, the force acting
on it is directed perpendicularly into the paper. A negatively charged
particle feels a force in the opposite direction.
Example (2)
An electron moving at 6.0x105 ms-1 passes perpendicularly through a
magnetic field of 2.0 ×10−2T. The charge of the electron is 6.1 ×10−19C.
what is the force on the electron

5 PHYSICS
EDEXCEL
 DR / MOSTA

Solution
F = BQV
−2 −19 5
F=2.0 ×10 ×1.6 × 10 × 6.0 ×10
−15
¿ 1.6 ×10 N (2 S . F )

The force between two current carrying conductors

If two current-carrying conductors are close
together, then each is in the magnetic field of the
other and therefore, each experiences a force. In
the opposite Fig X and Y are two infinitely long
parallel conductors carrying currents I1 and I2
respectively. The conductors are in vacuum and
their separation is a. the magnitude of the
magnetic flux density B, at any point P, on Y due
to the current in X is given by
I 1× I 2
F∝
a
By Fleming’s left-rule, the force in the wire is directed towards X as
shown; similar reasoning reveals that X is subject to a force of the
same magnitude. The force on X is in the opposite direction to that on
Y because the field along X is directed out of the paper.
Thus, the forces are such that the wires attract each other. It can be shown
at the current are in opposite directions the wires repel each other. Thus
Like currents attract, unlike currents repel
Couple on a coil in magnetic field
Current carrying coils in magnetic fields are essential components of
electric motors and meters of various kinds.
Consider a rectangular coil PQRS of N turns pivoted so that it can
rotate about a vertical axis YY’ which right angles to a uniform
magnetic field of flux density B, (fig a).
Let the normal to the plane of the coil make an angle  with the field, (fig
B) When the current I flows in the coil each side experiences a force (since
6 PHYSICS
EDEXCEL
 DR / MOSTA

all make some angle with B), acting perpendicularly to the plane
containing the side and the direction of the field. The forces on the top and
bottom (horizontal) sides are parallel to YY’ and for the current direction
shown, they lengthen the coil. the forces on the vertical sides, each of
length L, are equal and opposite and have value F where:
F=B× I × L × N
Whatever the position of the coil, its vertical sides are right angles to
B and so F remains constant. The forces continue a couple whose
torque C is given by
C = one force x perpendicular distance between lines of action of the
forces
= F × PT = F × b sin P Q T [b = width of coil]
= Fb sin θ
= BIA Nsin θ
C=BIAN θ

Where A area of face of coil = L x b

The couple cause angular acceleration of the
coil which rotates until its plane is
perpendicular to the field (θ =0) and then
C=0.
A circular coil carrying current I can be regarded as consisting of a
large number of tiny rectangular coils, each with current I flowing in
the same direction as in the circular coil.

7 PHYSICS
EDEXCEL
 DR / MOSTA

Angle bet. Coil

T = C = B I A N Cos θ & maq. field

Torque Couple
(Nm)

8 PHYSICS
EDEXCEL
 DR / MOSTA

*An electron in a magnetic field "change"

The force acting on a change in a mag. Field = Centripetal force
BQ V = mv
2

r
2
care∗mv 2
=m w r
r
2π
w=2 πf =
T
d
* r = 2 →cliamete

Force
2
mv 2
BIL=BQv=mg= =mw r
r

9 PHYSICS
EDEXCEL
 DR / MOSTA

* Prove that (c) couple = B I A N sin θ

(t) (torque)
∵ F=B I L sin θ

t=F × ⊥d

¿ B I L× ∝sin θ

= B I A sin θ for one turs

= B I A N sinθ for ( N ) turs

10 PHYSICS
EDEXCEL
 DR / MOSTA

Hall voltage:
The production of a potential difference across
a conductor carrying an electric current when a
magnetic field is applied in a direction
perpendicular to that of the current flow
Prove
F B=F E

Bq v=q E

VH
Bv=
d
I
V H =dBv V=
nAq
BI
V H =d
nAq
BI
V H=
ntq

Where:
FB is the magnetic force acting on the material (N)
FE is the electric force of the electrons passing through the conductor (N)
B is the magnetic flux density (T)
q is the charge of the current (C)
v is the velocity of the electrons passing through the conductor (ms-1)
d is the width of the conductor (m)
E is the electric field strength (NC-1)
T is thickness of the conductor (m)
n is the number of electrons in the conductor (m-3)
VH is the hall voltage (V)

11 PHYSICS
EDEXCEL
 DR / MOSTA

N.B
If you encounter a question like this: [Nov 20 V1, Q8(a)]
A slice of a conducting material has its face QRLK normal to a
uniform magnetic field of flux density B, as illustrated in Fig. 8.1.

Electrons enter the slice travelling perpendicular to face PQKJ.

(a) For the free electrons moving in the slice:
(i) state the direction of the force on an electron due to movement of
the electron in the magnetic field
(ii) identify the faces, using the letters on Fig. 8.1, between which a
potential difference is developed.

12 PHYSICS
EDEXCEL
 DR / MOSTA

You could answer with:

(i) Using Fleming’s left hand rule we can deduce
that the direction of the force is downwards
(ii) faces PQRS and JKLM
________________________________________________________
If you encounter a question like this: [Nov 20 V1, Q8(b)]
(b) Explain why the potential difference in (a)(ii) reaches a maximum
value.
You could answer with:
- (as charge separates) an electric field is created (between opposite
faces) (1 mark)
- (maximum value is reached when) electric force (on electron) is
equal and opposite to magnetic force (on electron) (1 mark)
If you encounter a question like this: [Nov 20 V1, Q8(c)]
The number of free electrons per unit volume in the slice of material is
1.3 × 1029m–3. The thickness PQ of the slice is 0.10mm. The magnetic
flux density B is 4.6 × 10–3T.
Calculate the potential difference across the slice for a current of
6.3×10–4A.

13 PHYSICS
EDEXCEL
 DR / MOSTA

You could answer with:

BI
V H=
ntq

¿ ( 4.6 ×10−3 ×6.3 × 10−4 ) / ( 1.3× 1029 × 0.10 ×10−3 ×1.60 × 10−19 ) (1 mark)
¿ 1.4 × 10
−12
V (1 mark)
If you encounter a question like this: [Nov 20 V1, Q8(d)]
The slice in (c) is a metal.
By reference to your answer in (c), suggest why Hall probes are
usually made using semiconductors rather than metals.
You could answer with:
- Semiconductors have a (much) smaller value for n (1 mark)
- VH for semiconductors is (much) larger so more easily measured (1
mark)
If you encounter a question like this: [June 19 V2, Q8(a)]
An electron is travelling in a vacuum at a speed of 3.4 × 107ms–1.
The electron enters a region of uniform magnetic field of flux density
3.2mT, as illustrated in Fig. 8.1.

14 PHYSICS
EDEXCEL
 DR / MOSTA

The initial direction of the electron is at an angle of 30° to the

direction of the magnetic field.
When the electron enters the magnetic field, the component of its
velocity vN normal to the direction of the magnetic field causes the
electron to begin to follow a circular path. Calculate:
(i) vN
(ii) the radius of this circular path
You could answer with:
(i) V N=3.4 ×107 ×sin 30 °=1.7 ×10 7 ms−1
2
mv mv
(ii) r
=bqv r=
bq (1 mark)
r =¿ (1 mark)
If you encounter a question like this: [June 19 V2, Q8(b)]
State the magnitude of the force, if any, on the electron in the
magnetic field due to the component of its velocity along the direction
of the field.
You could answer with:
Zero (because velocity has no effect on the force

15 PHYSICS
EDEXCEL
 DR / MOSTA

If you encounter a question like this: [June 19 V2, Q8(c)]

Use information from (a) and (b) to describe the resultant path of the
electron in the magnetic field.
You could answer with:
helix/coil
If you encounter a question like this: [Nov 19 V1, Q8(a)]
A long straight vertical wire carries a current I. The wire passes
through a horizontal card EFGH, as shown in Fig. 8.1 and Fig. 8.2.

On Fig. 8.2, draw the pattern of the magnetic field produced by the
current-carrying wire on the plane EFGH

(I) in x 8.4 same direction → Attraction and V.V.

F = B I L F = B I L
x y x x y x y y
If the current in the opposite direct → repulsion

16 PHYSICS
EDEXCEL
 DR / MOSTA

You could answer with:

Not equally spaced crowded at the center

* Mag. Flux (ϕ )
* Mag. Flux liskaje (Nϕ )
Nϕ
* Mag. Flux density (B) B=
A

17 PHYSICS
EDEXCEL