Chapter 28
28.58 The figure shows an end view of two long, parallel wires perpendicular to the xy -plane, each carrying
a current I but in opposite directions. (a) Copy the
~ field of each
diagram, and draw vectors to show the B
~
wire and the net B field at point P. (b) Derive the
~ at any point on
expression for the magnitude of B
the x- axis in terms of the x coordinate of the point.
~ (c) Graph the magnitude
What is the direction of B?
~ at points on the x- axis. (d) At what value of x
of B
~ a maximum? (e) What is the
is the magnitude of B
~
magnitude of B when x a?
(a) The diagram is shown on the right. Notice the four
equal angles all labeled by θ for which holds
1
q
a
a
.
cos θ = = √
2
r
a + x2
q
q
~1
(b) By symmetry the magnitudes of the two fields B
~ 2 are identical and equal to
and B
q
µ0 I
µ0
I
√
B1 = B2 =
=
2
2π r
2π a + x2
B1
r
r
Bnet
B2
2
Also by symmetry, the vertical components of the two fields cancel, i.e., By = B1y + B2y = 0,
~ net is horizontal and oriented to the right. Its magnitude is equal to the sum
so that the net field B
of the horizontal components of the two fields which are also identical:
Bnet = Bx = B1x + B2x = B1 cos θ + B2 cos θ = 2B1 cos θ = 2
i.e.,
Bnet =
µ0
I a
.
2
π a + x2
(d) At x = 0 so that
Bmax =
µ0 I
.
π a
(e)
Bmax ≈
µ0 I a
.
π x2
µ0
I
a
√
√
,
2π a2 + x2 a2 + x2
28.64 Calculate the magnetic field (magnitude
and direction) at a point P due to a current I =
12.0 A in the wire shown in the figure. Segment
BC is an arc of a circle with radius 30.0 cm,
and point P is at the center of curvature of the
arc. Segment DA is an arc of a circle with radius
20.0 cm, and point P is at its center of curvature.
Segments CD and AB are straight lines of length
10.0 cm each.
First notice that by the Biot-Savart law the magnetic field produced at P by any small segment
of of either of the two straight wire segments CD and AB is equl to zero. Thus the contribution to
the magnetic field at P comes only form the two arcs BC and AD.
Next, by the Biot-Savart law the direction of the magnetic field produced at P by any small
segment of length dl of the arc BC is perpendicular to the page and oriented out of it, while its
magnitude is
µ0 I dl
dBBC =
2 ,
4π rBC
where rBC = 0.300 m. Thus by integration the magnetic field produced by the entire segment BC
is also oriented out of the page and its magnitude is
BBC =
µ0 I LBC
,
2
4π rBC
where LBC is the length of the segment BC. By a simple proportion we conclude that
120◦
LBC
=
2πrBC
360◦
which implies LBC =
2
3
πrBC . Substituting this result into the expression for BBC we obtain
BBC =
µ0 I
= 8.38 × 10−6 T,
6 rBC
Repeating the same reasoning for the segment AD we conclude that at P it produces the field
oriented into the page and of magnitude
BAD =
µ0 I
= 1.26 × 10−5 T.
6 rAD
Thus, the net field is oriented into the page and its magnitude is
Bnet = BAD − BBC = 4.19 × 10−6 T.
28.80 A wide, long, insulating belt has a uniform positive charge per unit area σ on its upper
surface. Rollers at each end move the belt to the right at a constant speed v. Calculate the
magnitude and direction of the magnetic field produced by the moving belt at a point just above
its surface. (Hint: At points near the surface and far from its edges or ends, the moving belt can
be considered to be an infinite current sheet like that in Problem 28.69.)
current out of the page
current per width = l
B
current through shaded segment = l w
B
Amperian loop
w
The figure represents a cross sectional view of an infinite sheet perpendicular to the paper and
with the current per width λ directed out of the paper. This means that the current equal to λw
passes through the line in the sheet which is perpendicular to the current and of length w.
By symmetry at any point the magnetic field produced by the sheet is parallel to the sheet and
perpendicular to the direction of the current - this means that in the figure it is horizontal. In
addition above the sheet it is oriented to the left and below the sheet to the right, and a priori its
magnitude depends only on the distance from the sheet.
Introducing now a rectangular Amperian loop as shown in the figure, and applying Ampere’s
law we obtain
I
~ · d~l = 2B w = µ0 Ienc = µ0 λ w
B
loop
which implies
λ
.
2
Thus the magnitude of the field does not depend on the distance from the sheet. This is of course
the consequence of the fact that the sheet is assumed to be infinite.
The belt in the problem produces the same kind of field at the points sufficiently close to it and
far from its edges.
To find the equivalent current density λ observe that in time interval ∆t the amount of charge
that crosses the line of length w (which is perpendicular to the direction of the motion of the belt)
is equal to ∆q = σ∆A = σw∆L = σw v∆t where ∆L is the length traveled by the point of the belt
in the time interval ∆t. Thus the current ‘through the line’ of length w is ∆q/∆t = σw v which
means that the current per width is λ = σv. Substituting this into the above formula we obtain
that the magnetic field produced by the moving belt at a point just above its surface is
B = µ0
B = µ0
σv
.
2