This document discusses the Lorentz force experienced by a current-carrying conductor in a magnetic field. It states that the force (F) is proportional to the current (I), length of the conductor (l), and the sine of the angle between the conductor and the magnetic field (θ). It also provides examples of calculating the force on charges moving at various angles to a magnetic field, as well as the repulsive force between two parallel current-carrying conductors. Finally, it notes that a galvanometer can be converted to an ammeter by connecting a low resistance in parallel, and to a voltmeter by connecting a high resistance in series.

1. PHYSICS
FORCE ON A CURRENT
CARRYING CONDUCTOR IN A
MAGNETIC FIELD
Short revision series

2. Lorentz force of a current carrying
conductor
 consider charged particle q moving in a
where there is both magnetic and electric
field, the force exerted is F= Fe + Fm
 F = q (E + velocity x B)
 E = electric field
 B = bar magnet
 For a conductor in a field, the force F is
proportional to the current I flowing in the
conductor, lenght of the conductor and sine
angle the conductor makes with the field
 F α Il sinθ
 F = B I l sin θ B=magnetic flux density (Tesla)

3.  Magnetic flux = Ф = B A cos θ
 When a charged particle moves with a vel of
v in a magnetic field of flux density B
 Magnetic F = qv B sin θ
 If force is perpendicular F = qvB
 If force is parrallel F = 0

4. EG
 A conductor 1m long is carrying a current
flowing in a direction 450 with a field of 1T.
What is the current if the force experienced
is 1 N?

5.  F = B I l sin θ
 1 = 1 x I x sin 45
 1 = I sin 45
 I = 1/sin 45 = 1.175A

6. Eg 2
 A charge of +3μC moving with a velocity of
1x106 ms-1 in a flux of density 1T. What is
the force:
i) parallel to the field ii) perpendicular to the
field iii) at angle 450

7.  Parallel
 F = qv B sin θ = 3x10-6 x 1x106 x 1 sin 0 = 0
 Perpendicular
 F = qv B sin θ = 3x10-6 x 1x106 x 1 sin 90 = 3N
 At angle 450
 F = qv B sin θ = 3x10-6 x 1x106 x 1 sin45 =
2.55N

8. Force of 2 current carrying carrying
conductors
 When two current carrying conductors are
placed close the magnetic field produced by
one affects the other.
 F α I1 I2
l r *r = distance between them
F/l = μ0I1 I2
2πr
*μ0 = permeability of vacuum

9. Eg
 2 conductors carrying 2A and 3A currents
flowing in opposite direction seprated by 2m.
What is the repulsive force per unit lenght?
μ0 = 4π x 10-7 Am-1

10. F/l = μ0I1 I2
2πr
= 4π x 10-7 x 2 x 3 = 6 x 10-7 N
2π x 2

11. A LITTLE COMMON BUT TRICKY
QUESTION
 How do you convert a galvanometre to a
ammetre?
ANS: Connect a low resistance in parallel
 How do you convert a galvanometre to
voltmetre
ANS: Connect a high resistance in series.