This preview shows page 1. Sign up to view the full content.
Unformatted text preview: f the current, the palm
of the hand faces the direction of the magnetic force on the current.
The springs will stretch when the magnetic force exerted on the copper rod is downward,
toward the bottom of the page. Therefore, if you extend your right hand with your fingers
pointing out of the page and the palm of your hand facing the bottom of the page, your
thumb points lefttoright along the copper rod. Thus, the current flows left  to  right in the copper rod.
b. The downward magnetic force exerted on the copper rod is, according to Equation 21.3
F = ILB sin θ = (12 A)(0.85 m)(0.16 T) sin 90.0 ° = 1.6 N According to Equation 10.1, the force FxApplied required to stretch each spring is
FxApplied = kx , where k is the spring constant. Since there are two springs, we know that the magnetic force F exerted on the current must equal 2 FxApplied , so that F = 2 FxApplied = 2kx .
Solving for x, we find that
F
1.6 N
x=
=
= 1.1×10 –2 m
2k 2(75 N/m) 41. REASONING The following drawing shows a side view of the conducting rails and the
aluminum rod. Three forces act on the rod: (1) its weight mg, (2) the magnetic force F, and
the normal force FN. An application of RightHand Rule No. 1 shows that the magnetic
force is directed to the left, as shown in the drawing. Since the rod slides down the rails at a
constant velocity, its acceleration is zero. If we choose the xaxis to be along the rails,
Newton’s second law states that the net force along the xdirection is zero: ΣFx = max = 0.
Using the components of F and mg that are along the xaxis, Newton’s second law becomes 1162 MAGNETIC FORCES AND MAGNETIC FIELDS − F cos 30.0 ° + mg sin 30.0 ° = 0
14444 244444
4
3
∑ Fx The magnetic force is given by Equation 21.3 as F = ILB sin θ, where θ = 90.0° is the angle
between the magnetic field and the current. We can use these two relations to find the
current in the rod.
Current (out
of paper) FN Aluminum
rod
F
x mg
30.0 o
Conducting
rail SOLUTION Substituting the expression F = ILB sin 90.0° into Newton’s second law and
solving for the current I, we obtain
I= mg sin 30.0 °
=
LB sin 90.0 ° cos 30.0 ° b g 0
sin
b.20 kg g9.80 m / s h 30.0° =
c
1
sin
b.6 mg0.050 T g 90.0° cos 30.0°
b
2 14 A 42. REASONING According to Equation 21.4, the maximum torque is τmax = NIAB, where N
is the number of turns in the coil, I is the current, A = π r2 is the area of the circular coil, and
B is the magnitude of the magnetic field. We can apply the maximumtorque expression to
each coil, noting that τmax, N, and I are the same for each. SOLUTION Applying Equation 21.4 to each coil, we have τ max = NI π r12 B1
1442443 and τ max = NI π r22 B2 1442443 Coil 1 Coil 2 Dividing the expression for coil 2 by the expression for coil 1 gives τ max NIπ r22 B2
=
τ max NIπ r12 B1 or 1= r22 B2
r12 B1 Chapter 21 Problems 1163 Solving for r2, we obtain
r2 = r1 B1
B2 b = 5.0 cm T
g 0.18 T =
0.42 3.3 cm 43. REASONING According to Equation 21.4, the maximum torque is τmax = NIAB, where N is
the number of turns in the coil, I is the current, A = π r is the area of the circular coil, and B
is the magnitude of the magnetic field. Since the coil contains only one turn, the length L of
the wire is the circumference of the circle, so that L = 2π r or r = L/(2π). Since N, I, and B
are known we can solve for L.
2 SOLUTION According to Equation 21.4 and the fact that r = L/(2π), we have FL I B
Gπ J
HK
2
2 τ max = NIπ r 2 B = NIπ
Solving this result for L gives
L= 4 π τ max
NIB = c
h
1b
bg3.7 A g0.75 T g=
b 4 π 8.4 × 10 −4 N ⋅ m 0.062 m 44. REASONING The magnetic moment of a currentcarrying coil is discussed in
Section 21.6, where it is given as
Magnetic moment = NIA (1) In Equation (1), N is the number of turns in the coil, I is the current it carries, and A is its
area. Both coils in this problem are circular, so their areas are calculated from their radii via
A = π r2 .
SOLUTION Because the magnetic moments of the two coils are equal, Equation (1) yields N2 I2 A2 = N1I1A
1 (2) Substituting A = π r 2 into Equation (2) and solving for r2, we obtain ( ) ( 2
N2 I 2 π r2 = N1I1 π r12 ) or NI r22 = r12 1 1 N I 2 2 or r2 = r1 N1I1
N2 I 2 (3) 1164 MAGNETIC FORCES AND MAGNETIC FIELDS Therefore, the radius of the second coil is N1I1
= ( 0.088 m )
N2 I 2 r2 = r1 (140)( 4.2 A) = 0.053 m
(170)( 9.5A ) 45. REASONING According to Equation 21.4, the torque τ that the circular coil experiences is
τ = N I AB sinφ , where N is the number of turns, I is the current, A is the area of the circle, B
is the magnetic field strength, and φ is the angle between the normal to the coil and the
magnetic field. To use this expression, we need the area of the circle, which is π r 2 , where
r is the radius. We do not know the radius, but we know the length L of the wire, which
must equal the circumference of the single turn. Thus, L = 2π r , which can be solved for the
radius.
SOLUTION Using Equation 21.4 and the fact that the area A of a circle is A = π r 2 ,...
View
Full
Document
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

Click to edit the document details