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have that ( ) τ = N I AB sinφ = N I π r 2 B sinφ (1) Since the length of the wire is the circumference of the circle, or L = 2π r , it follows that the
L
radius of the circle is r =
. Substituting this result into Equation (1) gives
2π L 2 N I L2 B
τ = N I π r B sinφ = N I π B sinφ =
sinφ
4π 2π ( 2 ) The maximum torque τmax occurs when φ = 90.0°, so that τ max ( (1)( 4.30 A ) 7.00 ×10−2 m
N I L2 B
=
sin 90.0° =
4π
4π ) 2 ( 2.50 T ) sin 90.0° = 4.19 ×10−3 N ⋅ m 46. REASONING AND SOLUTION The torque is given by τ = NIAB sin φ.
a. The maximum torque occurs when φ = 90.0° (sin φ = 1). In this case we want the torque
to be 80.0% of the maximum value, so b NIAB sin φ = 0.800 NIAB sin 90.0 ° g so that φ = sin −1 e .800 j = 53.1°
0 Chapter 21 Problems b. The edgeon view of the coil at the right
shows the normal to the plane of the coil, the
magnetic field B, and the angle φ . 1165 B φ = 53.1° Coil Normal 47. SSM WWW REASONING The torque on the loop is given by Equation 21.4,
τ = NIABsin φ . From the drawing in the text, we see that the angle φ between the normal
to the plane of the loop and the magnetic field is 90° − 35° = 55° . The area of the loop is
0.70 m × 0.50 m = 0.35 m2.
SOLUTION
a. The magnitude of the net torque exerted on the loop is τ = NIAB sin φ = (75)(4.4 A)(0.35 m 2 )(1.8 T) sin 55° = 170 N ⋅ m
b. As discussed in the text, when a currentcarrying loop is placed in a magnetic field, the
loop tends to rotate such that its normal becomes aligned with the magnetic field. The
normal to the loop makes an angle of 55° with respect to the magnetic field. Since this angle
decreases as the loop rotates, the 35° angle increases . 48. REASONING When the wire is used to make a singleturn
1
square coil, each side of the square has a length of 4 L . When
the wire is used to make a twoturn coil, each side of the square
1
has a length of 8 L . The drawing shows these two options and
indicates that the total effective area of NA is greater for the
singleturn option. Hence, more torque is obtained by using the
singleturn option. SOLUTION According to Equation 21.4 the maximum torque
experienced by the coil is τmax = (NIAB) sin 90º = NIAB,
where N is the number of turns, I is the current, A is the area of
each turn, and B is the magnitude of the magnetic field.
Applying this expression to each option gives L/4 L/8 1166 MAGNETIC FORCES AND MAGNETIC FIELDS Single  turn τ max = NIAB = NI ( 1 L)
4 2 B 4 = (1)(1.7 A ) 1 (1.00 m ) Two  turn 2 ( 0.34 T ) = 0.036 N ⋅ m ( 0.34 T ) = 0.018 N ⋅ m τ max = NIAB = NI ( 1 L ) B
8
2 = ( 2 )(1.7 A ) 1 (1.00 m ) 8 2 49. REASONING The magnitude τ of the torque that acts on a currentcarrying coil placed in a
magnetic field is given by τ = NIAB sinφ (Equation 21.4), where N is the number of loops
in the coil (N = 1 in this problem), I is the current, A is the area of one loop, B is the
magnitude of the magnetic field (the same for each coil), and φ is the angle (the same for
each coil) between the normal to the coil and the magnetic field. Since we are given that the
torque for the square coil is the same as that for the circular coil, we can write (1) Isquare Asquare B sin φ 144424444
4
3
τ square = (1) I circle Acircle B sin φ
1444
2444
3
τ circle This relation can be used directly to find the ratio of the currents. SOLUTION Solving the equation above for the ratio of the currents yields
Isquare
I circle = Acircle
Asquare If the length of each wire is L, the length of each side of the square is
the square coil is Asquare = 1
4 L , and the area of 1
( 1 L ) ( 1 L ) = 16 L2 . The area of the circular coil is
4
4 Acircle = π r 2 , where r is the radius of the coil. Since the circumference (2π r) of the circular coil is equal to
the length L of the wire, we have 2π r = L, or r = L / ( 2π ) . Substituting this value for r into
the expression for the area of the circular coil gives Acircle = π [ L / ( 2π )] . Thus, the ratio of
the currents is
2 Isquare
I circle = Acircle
Asquare = 2 L 2π = 4 = 1.27
1 L2
π
16 π Chapter 21 Problems 1167 50. REASONING AND SOLUTION According to Equation 21.4, the maximum torque for a
single turn is τmax = IAB. When the length L of the wire is used to make the square, each
side of the square has a length L/4. The area of the square is Asquare = (L/4)2. For the
rectangle, since two sides have a length d, while the other two sides have a length 2d, it
follows that L = 6d, or d = L/6. The area is Arectangle = 2d 2 = 2(L/6)2. Using Equation 21.4 for
the square and the rectangle, we obtain τ square
τ rectangle = IAsquare B
IArectangle B Asquare = Arectangle L
b / 4g =
2b / 6g
L
2 = 2 1.13 51. REASONING The coil in the drawing is oriented such that the normal to the surface of the
coil is perpendicular to the magnetic field (φ = 90°). The magnetic torque is a maximum,
and Equation 21.4 gives its magnitude as τ = NIAB sin φ. In this expression N is the number
of loops in the coil, I is the current, A is the area of one loop, and B is th...
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