Physics For Scientists And Engineers 6E - part 226

SECTION 29.2 • Magnetic Force Acting on a Current-Carrying Conductor

901

perpendicular to and directed out of the page with a series of blue dots, which
represent the tips of arrows coming toward you (see Fig. 29.6a). In this case, we label
the field

out

. If

B is directed perpendicularly into the page, we use blue crosses, which

represent the feathered tails of arrows fired away from you, as in Figure 29.6b. In this
case, we label the field

, where the subscript “in” indicates “into the page.” The

same notation with crosses and dots is also used for other quantities that might be per-
pendicular to the page, such as forces and current directions.

One can demonstrate the magnetic force acting on a current-carrying conductor

by hanging a wire between the poles of a magnet, as shown in Figure 29.7a. For ease in
visualization, part of the horseshoe magnet in part (a) is removed to show the end face
of  the  south  pole  in  parts  (b),  (c),  and  (d)  of  Figure  29.7.  The  magnetic  field  is
directed  into  the  page  and  covers  the  region  within  the  shaded  squares.  When  the
current  in  the  wire  is  zero,  the  wire  remains  vertical,  as  shown  in  Figure  29.7b.
However, when the wire carries a current directed upward, as shown in Figure 29.7c,
the  wire  deflects  to  the  left.  If  we  reverse  the  current,  as  shown  in  Figure  29.7d,  the
wire deflects to the right.

Let us quantify this discussion by considering a straight segment of wire of length L

and cross-sectional area A, carrying a current I in a uniform magnetic field

B, as shown

in Figure 29.8. The magnetic force exerted on a charge q moving with a drift velocity
v

is q

B. To find the total force acting on the wire, we multiply the force q v

exerted on one charge by the number of charges in the segment. Because the volume
of the segment is AL, the number of charges in the segment is nAL, where n is the
number of charges per unit volume. Hence, the total magnetic force on the wire of
length L is

We can write this expression in a more convenient form by noting that, from Equation
27.4, the current in the wire is I " nqv

A. Therefore,

(29.3)

where

L is a vector that points in the direction of the current I and has a magnitude

equal to the length L of the segment. Note that this expression applies only to a
straight segment of wire in a uniform magnetic field.

L ! B

B)nAL

(a)

(b)

B out of page:

B into page:

Figure 29.6 (a) Magnetic field

lines coming out of the paper are

indicated by dots, representing the

tips of arrows coming outward.

(b) Magnetic field lines going into

the paper are indicated by crosses,

representing the feathers of arrows

going inward.

(b)

I = 0

(c)

(d)

(a)

Figure 29.7 (a) A wire suspended vertically between the poles of a magnet. (b) The

setup shown in part (a) as seen looking at the south pole of the magnet, so that the

magnetic field (blue crosses) is directed into the page. When there is no current in

the wire, it remains vertical. (c) When the current is upward, the wire deflects to the

left. (d) When the current is downward, the wire deflects to the right.

Figure 29.8 A segment of a

current-carrying wire in a magnetic

field B. The magnetic force

exerted on each charge making up

the current is q v

B and the net

force on the segment of length

L is I L ! B.

Force on a segment of

current-carrying wire in a

uniform magnetic field

Now consider an arbitrarily shaped wire segment of uniform cross section in

a magnetic field, as shown in Figure 29.9. It follows from Equation 29.3 that the
magnetic force exerted on a small segment of vector length d

s in the presence of a

field

B is

(29.4)

where d

is directed out of the page for the directions of

B and ds in Figure 29.9. We

can consider Equation 29.4 as an alternative definition of

B. That is, we can define the

magnetic field

B in terms of a measurable force exerted on a current element, where

the force is a maximum when

B is perpendicular to the element and zero when B is

parallel to the element.

To calculate the total force

acting on the wire shown in Figure 29.9, we integrate

Equation 29.4 over the length of the wire:

(29.5)

where a and b represent the end points of the wire. When this integration is carried
out, the magnitude of the magnetic field and the direction the field makes with the
vector d

s may differ at different points.

We now treat two interesting special cases involving Equation 29.5. In both cases,

the magnetic field is assumed to be uniform in magnitude and direction.

Case 1.

A curved wire carries a current I and is located in a uniform magnetic field

as shown in Figure 29.10a. Because the field is uniform, we can take

B outside the

integral in Equation 29.5, and we obtain

(29.6)

But the quantity d

s represents the vector sum of all the length elements from a to b.

From the law of vector addition, the sum equals the vector

L(, directed from a to b.

Therefore, Equation 29.6 reduces to

(29.7)

From this we conclude that

the magnetic force on a curved current-carrying wire

in a uniform magnetic field is equal to that on a straight wire connecting the
end points and carrying the same current.

L( ! B

s ! B

I d

s ! B

902

C H A P T E R 2 9 • Magnetic Fields

Figure 29.9 A wire segment of

arbitrary shape carrying a current I

in a magnetic field B experiences a

magnetic force. The magnetic

force on any segment ds is I ds ! B

and is directed out of the page. You

should use the right-hand rule to

confirm this force direction.

(b)

d s

′

(a)

Figure 29.10 (a) A curved wire carrying a current I in a uniform magnetic field.

The total magnetic force acting on the wire is equivalent to the force on a straight wire

of length L( running between the ends of the curved wire. (b) A current-carrying loop

of arbitrary shape in a uniform magnetic field. The net magnetic force on the loop

is zero.

SECTION 29.2 • Magnetic Force Acting on a Current-Carrying Conductor

903

Case 2.

An arbitrarily shaped closed loop carrying a current I is placed in a uniform

magnetic field, as shown in Figure 29.10b. We can again express the magnetic force
acting on the loop in the form of Equation 29.6, but this time we must take the vector
sum of the length elements d

s over the entire loop:

Because the set of length elements forms a closed polygon, the vector sum must be
zero. This follows from the procedure for adding vectors by the graphical method.
Because

% ds " 0, we conclude that F

0; that is,

the net magnetic force acting on

any closed current loop in a uniform magnetic field is zero.

Quick Quiz 29.4

The four wires shown in Figure 29.11 all carry the same

current from point A to point B through the same magnetic field. In all four parts of
the figure, the points A and B are 10 cm apart. Rank the wires according to the magni-
tude of the magnetic force exerted on them, from greatest to least.

Quick Quiz 29.5

A wire carries current in the plane of this paper toward

the top of the page. The wire experiences a magnetic force toward the right edge of
the page. The direction of the magnetic field causing this force is (a) in the plane
of the page and toward the left edge, (b) in the plane of the page and toward the
bottom edge, (c) upward out of the page, (d) downward into the page.

(a)

(b)

(c)

(d)

Figure 29.11 (Quick Quiz 29.4) Which wire experiences the greatest magnetic force?

Example 29.2 Force on a Semicircular Conductor

A  wire  bent  into  a  semicircle  of  radius  R forms  a  closed
circuit  and  carries  a  current  I.  The  wire  lies  in  the  xy
plane,  and  a  uniform  magnetic  field  is  directed  along
the positive  y axis,  as  shown  in  Figure  29.12.  Find  the
magnitude  and  direction  of  the  magnetic  force  acting

on the straight portion of the wire and on the curved
portion.

Solution The magnetic force

acting on the straight por-

tion has a magnitude F

ILB " 2IRB because L " 2R and

29.3 Torque on a Current Loop in a Uniform

Magnetic Field

In the preceding section, we showed how a magnetic force is exerted on a current-
carrying conductor placed in a magnetic field. With this as a starting point, we now
show that a torque is exerted on a current loop placed in a magnetic field. The results
of this analysis will be of great value when we discuss motors in Chapter 31.

Consider a rectangular loop carrying a current I in the presence of a uniform mag-

netic field directed parallel to the plane of the loop, as shown in Figure 29.13a. No
magnetic forces act on sides ! and " because these wires are parallel to the field;
hence,

L ! B " 0 for these sides. However, magnetic forces do act on sides # and $

because these sides are oriented perpendicular to the field. The magnitude of these
forces is, from Equation 29.3,

The direction of

, the magnetic force exerted on wire #, is out of the page in the view

shown in Figure 29.13a, and that of

, the magnetic force exerted on wire $, is into the

page in the same view. If we view the loop from side " and sight along sides # and $,
we see the view shown in Figure 29.13b, and the two magnetic forces

and

are di-

rected as shown. Note that the two forces point in opposite directions but are not di-
rected along the same line of action. If the loop is pivoted so that it can rotate about
point O, these two forces produce about O a torque that rotates the loop clockwise. The
magnitude of this torque )

max

where the moment arm about O is b/2 for each force. Because the area enclosed by the
loop is A " ab, we can express the maximum torque as

(29.8)

This maximum-torque result is valid only when the magnetic field is parallel to the plane
of the loop. The sense of the rotation is clockwise when viewed from side ", as indicated
in Figure 29.13b. If the current direction were reversed, the force directions would also
reverse, and the rotational tendency would be counterclockwise.

)

max

IAB

)

max

(IaB)

IabB

IaB

904

C H A P T E R 2 9 • Magnetic Fields

the wire is oriented perpendicular to

B. The direction of F

is out of the page based on the right-hand rule for the cross
product

L ! B.

To find the magnetic force

acting on the curved part,

we  use  the  results  of  Case  1.  The  magnetic  force  on  the
curved  portion  is  the  same  as  that  on  a  straight  wire  of
length  2R carrying  current  I to  the  left.  Thus,  F

ILB "

2IRB. The direction of

is into the page based on the

right-hand rule for the cross product

L ! B.

Because the wire lies in the xy plane, the two forces on

the loop can be expressed as

The net magnetic force on the loop is

Note that this is consistent with Case 2, because the wire
forms a closed loop in a uniform magnetic field.

F " F

2IR

B ˆ

k $ 2IR

B ˆ

k " 0

2IR

B ˆ

2IR

B ˆ

Figure 29.12 (Example 29.2) The net magnetic force acting

on a closed current loop in a uniform magnetic field is zero. In

the setup shown here, the magnetic force on the straight

portion of the loop is 2IRB and directed out of the page, and

the magnetic force on the curved portion is 2IRB directed into

the page.

(a)

(b)

Figure 29.13 (a) Overhead view

of a rectangular current loop in a

uniform magnetic field. No mag-

netic forces are acting on sides !

and " because these sides are paral-

lel to B. Forces are acting on sides
#

and $, however. (b) Edge view of

the loop sighting down sides # and
$

shows that the magnetic forces F

and F

exerted on these sides create

a torque that tends to twist the loop

clockwise. The purple dot in the left

circle represents current in wire #

coming toward you; the purple cross

in the right circle represents current

in wire $ moving away from you.

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