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Unformatted text preview: EASONING Since the two wires are next to each other, the net magnetic field is
everywhere parallel to ∆l in Figure 21.37. Moreover, the net magnetic field B has the same
magnitude B at each point along the circular path, because each point is at the same distance
from the wires. Thus, in Ampère's law (Equation 21.8), B = B , I = I1 + I 2 , and we have ΣB ∆l = B ( Σ∆l ) = µ0 ( I1 + I 2 )
But Σ∆l is just the circumference (2πr) of the circle, so Ampère's law becomes B ( 2π r ) = µ0 ( I1 + I 2 )
This expression can be solved for B. SOLUTION
a. When the currents are in the same direction, we find that ( ) –7
µ0 ( I1 + I 2 ) 4π × 10 T ⋅ m/A ( 28 A + 12 A )
B=
=
= 1.1 × 10 –5 T
2π r
2π ( 0.72 m ) Chapter 21 Problems 1179 b. When the currents have opposite directions, a similar calculation shows that ( ) –7
µ0 ( I1 – I 2 ) 4π × 10 T ⋅ m/A ( 28 A–12 A )
B=
=
= 4.4 × 10 –6 T
2π r
2π ( 0.72 m ) 69. SSM REASONING AND SOLUTION Ampère's law in the form of Equation 21.8
indicates that ΣB∆l = µ0 I . Since the magnetic field is everywhere perpendicular to the
plane of the paper, it is everywhere perpendicular to the circular path and has no component
B that is parallel to the circular path. Therefore, Ampère's law reduces to
ΣB∆l = 0 = µ0 I , so that the net current passing through the circular surface is zero . 70. REASONING Both parts of this problem can be solved using Ampère’s law with the
circular closed paths suggested in the Hint given with the problem statement. The circular
closed paths are used because of the symmetry in the way the current is distributed on the
copper cylinder. SOLUTION
a. An endon view of the copper cylinder is a
circle, as the drawing at the right shows. The
dots around the circle represent the current
coming out of the paper toward you. The
larger dashed circle of radius r is the closed
path used in Ampère's law and is centered on
the axis of the cylinder. Equation 21.8 gives
Ampère's law as ΣB∆l = µ0 I . Because of the
symmetry of the arrangement in the drawing,
we have B = B for all ∆l on the circular path,
so that Ampère's law becomes Circular path,
radius = r Copper
cylinder ΣB∆l = B ( Σ∆l ) = µ0 I In this result, I is the net current through the circular surface bounded by the dashed path. In
other words, it is the current I in the copper tube. Furthermore, Σ∆l is the circumference of
the circle, so we find that
µI
B ( Σ∆l ) = B ( 2π r ) = µ0 I or
B= 0
2π r 1180 MAGNETIC FORCES AND MAGNETIC FIELDS Circular path,
radius = r b. The setup here is similar to that in part a, except that
the smaller dashed circle of radius r is now the closed
path used in Ampère's law (see the drawing at the
right). With this change, the derivation then proceeds
exactly as in part a. Now, however, there is no current
through the circular surface bounded by the dashed
path, because all of the current is outside the path.
Therefore, I = 0 A, and B= Copper
cylinder µ0I
= 0T
2π r 71. REASONING AND SOLUTION The drawing at the
right shows an endon view of the solid cylinder. The
dots represent the current in the cylinder coming out
of the paper toward you. The dashed circle of radius r
is the closed path used in Ampère's law and is
centered on the axis of the cylinder. Equation 21.8
gives Ampère's law as ΣB∆l = µ0 I . Because of the
symmetry of the arrangement in the drawing, we have
B = B for all ∆l on the circular path, so that Ampère's law becomes Solid
cylinder
Circular path,
radius = r ΣB∆l = B ( Σ∆l ) = µ0 I In this result, Σ∆l = 2π r, the circumference of the circle. The current I is the part of the
total current that comes through the area π r 2 bounded by the dashed path. We can
calculate this current by using the current per unit crosssectional area of the solid cylinder.
This current per unit area is called the current density. The current I is the current density
times the area π r 2 :
I r2
I I = 02 π r2 = 0 2
R
π
4 R4
123 ( ) current
density Thus, Ampère's law becomes
B ( Σ∆l ) = µ0 I or I r2 B ( 2π r ) = µ0 0 2 R or B= µ0 I 0 r
2π R 2 Chapter 21 Problems 72. REASONING The magnitude B of the magnetic field is given by B = 1181 F
q0 v sin θ (Equation 21.1), and we will apply this expression directly to obtain B.
SOLUTION The charge q0 = −8.3 ×10−6 C travels with a speed v = 7.4 × 106 m/s at an angle of θ = 52° with respect to a magnetic field of magnitude B and experiences a force of
magnitude F = 5.4 × 10−3 N. According to Equation 21.1, the field magnitude is
B= F
5.4 ×10−3 N
=
= 1.1× 10−4 T
q0 v sin θ −8.3 × 10−6 C 7.4 × 106 m/s sin 52° ( ) Note in particular that it is only the magnitude q0 of the charge that appears in this
calculation. The algebraic sign of the charge does not affect the result. 73. REASONING AND SOLUTION The magnitude B of the magnetic field at a distance r
from a long straight wire carrying a current I is B = µ0I/(2π r). Thus, the distance is c b
hg
h 4...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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