A current-carrying conductor placed in an external magnetic field experiences a force. This force is described by the Right-Hand Rule #3. Two parallel current-carrying conductors also experience a force of attraction or repulsion depending on whether the currents are in the same or opposite directions. The magnitude of the force between conductors can be calculated using the given equation. An example calculation is also provided to find the force on a conductor using the magnetic field strength, current, and length of the conductor in the field. Practice questions are listed at the end to test understanding of these concepts.

1. Lesson 3
RHR # 3
Nelson Reference Pages: 563-564Nelson Reference Pages: 563-564
(Not all parts are covered in Nelson TB)(Not all parts are covered in Nelson TB)
{McGraw-Hill Reference: 700-702, 705-709}{McGraw-Hill Reference: 700-702, 705-709}

2.  When a current carryingWhen a current carrying
conductor isconductor is ⊥⊥ to anto an
external magnetic field,external magnetic field,
there will be a force actingthere will be a force acting
on the conductor. Theon the conductor. The
direction of the force isdirection of the force is
found by using RHR #3 – afound by using RHR #3 – a
palm rule.palm rule. If the conductor
is // to the field, there is, there is no
force acting on it.force acting on it.
N
S
F
U S E R H R # 3 T O F IN D
T H E D IR E C T IO N O F F
RHR #3: Thumb is in the direction of the current.
Fingers in the direction of the external magnetic
field, and out of the palm is the direction of the
force acting on the conductor.

3.  If two current carrying conductors are // to each other,If two current carrying conductors are // to each other,
they will experience either a force of attraction orthey will experience either a force of attraction or
repulsion. The magnitude of this force can be foundrepulsion. The magnitude of this force can be found
from the following equation:from the following equation: F = k IAIBL d-1
, here, here k = 2.0
x 10-7
N/A2
,, II is the current in each conductoris the current in each conductor L is theis the
common length in metres andin metres and d is the separation ofis the separation of
the conductors in metres. RHR 1 and RHR 3 can bethe conductors in metres. RHR 1 and RHR 3 can be
used to determine the direction of the force.used to determine the direction of the force.
X X
F F
IF C U R R E N T S A R E IN T H E S A M E
D IR E C T IO N , T H E R E W IL L B E A F O R C E O F
A T T R A C T IO N .
N O T E , F IE L D L IN E S D O N O T C R O S S !

4. The Motor Force (see diagram first slide):
 The force acting on the conductor that isThe force acting on the conductor that is ⊥⊥ in anin an
external magnetic field is given by:external magnetic field is given by: F = B⊥IL,
herehere BB⊥⊥ is the magnetic field strength measuredis the magnetic field strength measured
in Tesla (T),in Tesla (T), II is in Amperes (A), andis in Amperes (A), and L is theis the
length of the conductor in the magnetic field inlength of the conductor in the magnetic field in
metres. { If a number of coils (metres. { If a number of coils (n) of length) of length l areare
in a magnetic field, thenin a magnetic field, then L = nl }}
 1 T = 1 N/(A m)
Although Oersted is given credit for the discoveryAlthough Oersted is given credit for the discovery
of electromagnetism, it wasof electromagnetism, it was Andre Marie Ampere
that developed most of the theory including thethat developed most of the theory including the
RHR’s.RHR’s.

5.  Example: A 20 cm conductor isExample: A 20 cm conductor is ⊥⊥ to ato a
magnetic field of strength 2.0 T. If themagnetic field of strength 2.0 T. If the
current in the conductor is 1.5 A, determinecurrent in the conductor is 1.5 A, determine
the magnitude of the force acting on thethe magnitude of the force acting on the
conductor.conductor.
 F = B⊥IL = 2.0 T X 1.5 A x 0.20 m = 0.60 N

6. Practice Questions
Nelson TB:
Page 562 # 3Page 562 # 3
Page 566 # 1, 2Page 566 # 1, 2
Page 582 # 31Page 582 # 31
Workbook
Page 63 # 1, 2, 3 a,c (ignore b), 4 i, ii, 5Page 63 # 1, 2, 3 a,c (ignore b), 4 i, ii, 5
Continued on next page…

7. 1.1. A magnetic field has a strength of 1.2 T into theA magnetic field has a strength of 1.2 T into the
page. A current of 7.5 A flows vertically upwardpage. A current of 7.5 A flows vertically upward
through a conductor that has 0.080 m inside thethrough a conductor that has 0.080 m inside the
field. Find the force that the field exerts on thefield. Find the force that the field exerts on the
conductor. [conductor. [Ans 0.72 N [left]]0.72 N [left]] (McGraw page 710 #1)(McGraw page 710 #1)
(McGraw page 715#30)(McGraw page 715#30)
2.2. Two parallel conductors, which may beTwo parallel conductors, which may be
considered close together, carry currents inconsidered close together, carry currents in
opposite directions. Describe what would happenopposite directions. Describe what would happen
to the magnitude of the force if:to the magnitude of the force if:
a.a. The current in one conductor is doubled (2 x Incr)The current in one conductor is doubled (2 x Incr)
b.b. The current in both conductors is tripled.(9 x Incr)The current in both conductors is tripled.(9 x Incr)
c.c. The distance between the conductors is halved.The distance between the conductors is halved.
(2 x Incr)(2 x Incr)

8.  Other questions from McGraw-HillOther questions from McGraw-Hill
 Page 710 #1-4Page 710 #1-4
 Page 715 #30Page 715 #30