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Unformatted text preview: e magnitude of the
magnetic field. The torque from the brake balances this magnetic torque. The brake torque is
τbrake = Fbraker, where Fbrake is the brake force, and r is the radius of the shaft and also the
lever arm. The maximum value for the brake force available from static friction is
Fbrake = µsFN (Equation 4.7), where FN is the normal force pressing the brake shoe against
the shaft. The maximum brake torque, then, is τbrake = µsFNr. By setting τbrake = τmax, we
will be able to determine the magnitude of the normal force.
SOLUTION Setting the torque produced by the brake equal to the maximum torque
produced by the coil gives τ brake = τ or µs FN r = NIAB sin φ ( ) –3 2
NIAB sin φ ( 410 )( 0.26 A ) 3.1 × 10 m ( 0.23 T ) sin 90°
FN =
=
= 8.3 N
µs r
( 0.76 )( 0.012 m ) 52. REASONING
The magnetic moment of the rotating charge can be found from the
expression Magnetic moment = NIA , as discussed in Section 21.6. For this situation, N = 1.
Thus, we need to find the current and the area for the rotating charge. This can be done by
resorting to first principles.
SOLUTION The current for the rotating charge is, by definition (see Equation 20.1),
I = ∆q / ∆t , where ∆q is the amount of charge that passes by a given point during a time
interval ∆t. Since the charge passes by once per revolution, we can find the current by
dividing the total rotating charge by the period T of revolution. 1168 MAGNETIC FORCES AND MAGNETIC FIELDS I= ∆q
∆q
ω ∆q
(150 rad/s)(4.0 ×10 –6 C)
=
=
=
= 9.5 ×10 –5 A
T
2π / ω
2π
2π The area of the rotating charge is A = π r 2 = π (0.20 m)2 = 0.13 m 2
Therefore, the magnetic moment is
Magnetic moment = NIA = (1)(9.5 × 10 –5 A)(0.13 m 2 ) = 1.2 × 10 –5 A ⋅ m 2 53. SSM REASONING AND SOLUTION
a. In Figure 21.26a the magnetic field that exists at the location of each wire points upward.
Since the current in each wire is the same, the fields at the locations of the wires also have
the same magnitudes. Therefore, a single external field that points downward will cancel
the mutual repulsion of the wires, if this external field has a magnitude that equals that of
the field produced by either wire.
b. Equation 21.5 gives the magnitude of the field produced by a long straight wire. The
external field must have this magnitude: ( ) 4π × 10 –7 T ⋅ m/A ( 25 A )
µ0 I
B=
=
= 3.1 × 10 –4 T
2π r
2π ( 0.016 m )
54. REASONING The torque τ exerted on a coil with a current Icoil is given by
τ = NI coil AB sin φ (Equation 21.4), where N is the number of turns in the coil, A is its area,
B is the magnitude of the external magnetic field causing the torque, and φ is the angle
between the normal to the surface of the coil and the direction of the external magnetic field.
The external magnetic field B, in this case, is the magnetic field of the solenoid. We will use
B = µ0 nI (Equation 21.7) to determine the magnetic field, where µ0 = 4π ×10−7 T ⋅ m/A is
the permeability of free space, n is the number of turns per meter of length of the solenoid,
and I is the current in the solenoid. The magnetic field in the interior of a solenoid is parallel
to the solenoid’s axis. Because the normal to the surface of the coil is perpendicular to the
axis of the solenoid, the angle φ is equal to 90.0°.
SOLUTION Substituting the expression B = µ0 nI (Equation 21.7) for the magnetic field of the solenoid into τ = NI coil AB sin φ (Equation 21.4), we obtain τ = NI coil AB sin φ = NI coil A ( µ0 nI ) sin φ = µ0 nNAI coil I sin φ Chapter 21 Problems 1169 Therefore, the torque exerted on the coil is τ = ( 4π ×10−7 T ⋅ m/A ) (1400 m −1 ) ( 50 ) (1.2 ×10−3 m 2 ) ( 0.50 A )( 3.5 A ) sin 90.0o
= 1.8 ×10−4 N ⋅ m 55. SSM REASONING The magnitude B of the magnetic field in the interior of a solenoid
that has a length much greater than its diameter is given by B = µ0 nI (Equation 21.7),
where µ0 = 4π ×10−7 T ⋅ m/A is the permeability of free space, n is the number of turns per
meter of the solenoid’s length, and I is the current in the wire of the solenoid. Since B and I
are given, we can solve Equation 21.7 for n. SOLUTION Solving Equation 21.7 for n, we find that the number of turns per meter of
length is
B
7.0 T
n=
=
= 2.8 ×104 turns/m
−7
2
µ0 I
4π × 10 T ⋅ m/A 2.0 × 10 A ( )( ) 56. REASONING AND SOLUTION The magnetic field at the center of a current loop of
radius R is given by B = µ 0I/(2R), so that
R= µ0 I
2B = ( 4π × 10−7 T ⋅ m/A ) (12 A )
2 (1.8 × 10−4 T ) = 4.2 × 10−2 m 57. SSM REASONING The magnitude of the magnetic field at the center of a circular loop
of current is given by Equation 21.6 as B = Nµ0I/(2R), where N is the number of turns, µ0 is
the permeability of free space, I is the current, and R is the radius of the loop. The field is
perpendicular to the plane of the loop. Magnetic fields are vectors, and here we have two
fields, each perpendicular to the plane of the loop producing it. Therefore, the two field
vectors are perpendicular, and we must...
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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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