21.6 The Torque on a Current-Carrying Coil
Put a rectangular loop of current I and length (height) L, and width w in a uniform magnetic
field B. The loop is mounted such that it is free to rotate about a vertical axis through its
center. We will consider the forces on each segment and the resulting torque from each.
Using RHR-1: The force on segment 3 points down, and that on segment 4 points up.
F3 and F4 are also equal in magnitude and cancel one another.
The magnitudes F3 = F4 = IwBsin(90°-φ) = IwBcosφ also change with the rotation angle φ
But both F3 and F4 are directed parallel to the axis, and results in no torque.
Top
view
3
3
4
φ is the angle between
the “normal” to the loop
and the magnetic field
1
21.6 The Torque on a Current-Carrying Coil
Looking at segments 1 and 2 which have the current running vertically.
By RHR-1, force F1 on segment 1 (current up) points into the page, for all values of φ. Also
by RHR-1, force F2 on segment 1 (current down) points out of the page. They cancel each
other to yield no net force on the loop.
However, F1 and F2 both tend to turn the loop in the clockwise sense (as seen in the top
view). The torques from the two forces are each
 w
τ 1, 2 = (F1, 2 )  sin φ
2
τ = τ 1 + τ 2 = Fw sin φ
Top view
F = ILB sin θ = ILB
since θ = 90°
2
1
φ is the angle between
the “normal” to the loop
and the magnetic field
2
21.6 The Torque on a Current-Carrying Coil
τ = τ 1 + τ 2 = Fw sin φ
The torque τ is maximum when the normal of the
loop is perpendicular to the magnetic field, and
zero when the normal is parallel to the field.
F = ILB sin θ = ILB
since θ = 90°
The torque tends to cause the loop normal to
become aligned to the field, just like on a bar
magnet. Current loop = magnetic dipole
Net torque = τ = ILB (w sin φ ) = IAB sin φ
Top view
A = Lw =
area of loop
magnetic
moment m

τ = NIA B sin φ
number of turns of wire
3
21.6 The Torque on a Current-Carrying Coil
Example The Torque Exerted on a Current-Carrying Coil
A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a
current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T.
(a) Determine the magnetic moment of the coil.
(b) Find the maximum torque that the magnetic field can exert on the coil.
4
21.6 The Torque on a Current-Carrying Coil
Example The Torque Exerted on a Current-Carrying Coil
A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and
contains a current of 0.045 A. The coil is placed in a uniform magnetic field of
magnitude 0.15 T.
(a) Determine the magnetic moment of the coil.
(b) Find the maximum torque that the magnetic field can exert on the coil.
magnetic
moment

(a) m = NIA = (100 )(0.045 A )(2.0 × 10 − 4 m 2 ) = 9.0 × 10 − 4 A ⋅ m 2
magnetic
moment

−4
2

−4
(b) τ = NIA B sin φ = 9.0 × 10 A ⋅ m (0.15 T ) sin 90 = 1.4 × 10 N ⋅ m
(
)
5
21.6 The Torque on a Current-Carrying Coil
Application: The basic components of
a dc motor.
The brushes switches the
direction of the current so
that the torque is always in
the same direction 
continuous rotation
6
22.1 Induced Emf and Induced Current
Chapter 22
• Electromagnetic Induction
In Class Demo
It is the changing field that produces the current.
The current in the coil is called the induced current because it is brought about by a
changing magnetic field.
Since a source emf (such as a voltage source) is needed to produce a current, the
coil behaves as if it were a source of emf. This emf is known as the induced emf.
7
22.1 Induced Emf and Induced Current
An emf can be induced by changing
the area of a coil in a constant
magnetic field
In each example, both an emf and a
current are induced because the coil is
part of a complete circuit. If the circuit
were open, there would be no induced
current, but there would still be an
induced emf.
I
The phenomena of producing an
induced emf with the aid of a
magnetic field is called
electromagnetic induction.
The induced emf depends on:
the magnitude of the magnetic field
the orientation of the magnetic field
the speed of the magnet
Rate of change of the area of the coil or the
speed of the coil
size of the coil
8
22.2 Motional Emf
THE EMF INDUCED IN A MOVING CONDUCTOR
Conducting
bar not
touching the
plates

F
+
≡
(is equivalent to)
Each charge within the conductor is
moving and experiences a magnetic
force
F = qvB

F
+
∆V = EL
= vBL
The magnetic force is in fact exactly
equivalent to an electric force exerted by a
parallel plate capacitor moving with the rod.
F = qE , E = vB, ∆V = EL = vBL
The action, on a moving charge, of a magnetic field generated by a magnet at rest
is exactly equivalent to that of an electric field generated in the rest frame (the
frame of reference in which the object in question is at rest) of that charge.
(not in the scope of this class)
9
22.2 Motional Emf
From a different point of view, the bar behaves as if it has a built-in
battery (or DC voltage source). As illustrated on the right side of the
figure, the battery will also cause charges to accumulate at the ends.

F
+
≡
(is equivalent to)
Motional emf when v, B,
and L are mutually
perpendicular
E = vBL
+
−
E

v
The direction of the EMF
(− to the + side of the
battery) is in the same
direction as the magnetic
force given by RHR-1
10
22.2 Motional Emf
The moving bar is in fact exactly equivalent to a battery, that one can
construct a circuit with it to power a light bulb. The diagrams below
show such a circuit made using conducting rails over which the bar
slides while making full electrical contact to form a closed circuit.
≡
I
(is equivalent to)
E = vBL
11
22.2 Motional Emf
Example : Operating a Light Bulb
with Motional Emf
In the figure, the conducting rod is moving with a speed of
5.0m/s perpendicular to a 0.80T magnetic field.
The rod has a length of 1.6m and a negligible electrical
resistance.
The rails also have a negligible electrical resistance. The
light bulb has a resistance of 96 ohms. Find
(a) the emf produced by the rod and
(b) the current induced in the circuit.
12
22.2 Motional Emf
Example : Operating a Light Bulb with Motional Emf
In the figure, the conducting rod is
moving with a speed of 5.0m/s
perpendicular to a 0.80T magnetic field.
The rod has a length of 1.6m and a
negligible electrical resistance.
The rails also have a negligible
electrical resistance. The light bulb has
a resistance of 96 ohms. Find
(a) the emf produced by the rod and
(b) the current induced in the circuit.
(a) E = vBL = (5.0 m s )(0.80 T )(1.6 m ) = 6.4 V
(b) I =
E 6.4 V
=
= 0.067 A
R 96Ω
13
22.2 Motional Emf
But where does the power and energy
come from to make the light bulb shine?
That hand has to work against the
magnetic force exerted by the field on the
induced current.
RHR-1 gives a magnetic force to the left.


Fhand = − FB ,
In order to keep the rod moving at
constant velocity, the force the hand
exerts on the rod must balance the
magnetic force on the current:


Fhand = FB = ILB
The opposite direction of the force from
the previous page, shown in this figure
below, would violate the principle of
conservation of energy.
A magnetic force produced in the
same direction as the original velocity
of the bar would then accelerate the
bar without any external work: bar
will accelerate and the force would
increase.
THIS DOES NOT HAPPEN
14