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Unformatted text preview: 20.7 a) The magnetic force on a charge q is ﬁquXﬁ. The magnitude of the force on the electron is F =[— el‘UBsinB = (1.502 x 10"” C)(5.G x 105 m/s)(6.0 x 10“2 Tlsin 150° m: F:2.4>< 10—14N. ge, the force on it is directed opposite to 17 X 1—3:. Here, the vector product
direction of i" X B is perpendicular and out of the page, hence the force
is perpendicularly into the page. b) Find the magnitude of the acceleration by applying Newton’s second to the electron, using the magni—
tudes of the vectors,  F 2.4le—14N
F:ma 2} {1:}; = = l 16 2.
ammo31kg 26x10 m/s 20.27 a) Consider the square loop. The magnitude of the torque 0n the loop is T I [fix ﬁg =pBSin€9 : IABsinB :— (150 A)(20.0 x 104 m)2(1.0 x 10W3 T}sin60° 2:» 7” = 5.2 x 10“4 Nm. b) An equilateral triangle with sides of length [5' has area 1 l . o
A z %(ba5e)(height) 2 §(E){€ sin 60°) = 532 8111 60 .
The torque on this loop has magnitude ' 1 . o _ . o
T : IABsin§ 2:, 5.2 x 10—4 Nm = (15.0 A) (522 511160 ) (1.0 >010 3 T)sm6{) 2A; €=30x1(}“2m. c) A circle of radius R has area A 2 TFR2.
The torque on this loop has magnitude
r = IABsine m; 5.2 x 10—4 Nrn = (15.0 A)(7rR2)(1.O ><10“3 T) sin60° E . 2> R211x10”2m. 7'60 CHAPTER 20. MAGNETIC FORCES AND THE MAGNETIC FIELD 20.33 The torque on'the loop is 7" = ,E x g, so the magnitude of the torque is 7' = p.13 sind. The maximum magnitude is when the angle 6 between [:5 and E is 90°,so 'r =szinl9 :pB. The magnitude of the magnetic dipole moment is the product of the current I, the area A of the coil, and
the number n of coils, e":an 2; rnnmg 2; A: 7' 20{3Nm _._v— M = D —2 '2
MB 100(25.0A)(0:200 T) 400 X 10 m ' 7 20.48 a) The distance d of point P from either wire is
d = (0.25 In) sin 300 = 0.13 m. g ance d from an inﬁnite wire is given in Table 20.1 on page 914 of
the text ' w#021 47rd. The magnetic ﬁeid at P from the 30 A current is directed perpendicular to and out of the page, While the ﬁeld at P from the 10 A current is directed perpendicular and into the page. The total ﬁeld at P is the
vector sum of the two ﬁelds, which has magnitude —7 . A *7  A 2
B:WMW2MMO—ST c1,“ 10—5T rammesr,
0.13m 0.13m and is directed perpendicular to and out of the page.
b) The distance d of point 8 from either wire is d = (0.25 In) sin 60" = 8.22 m. At point 8, the magnetic ﬁeld from the 30 A current at P is directed perpendicular to and out of the page, Whilethe ﬁeld from the it} A current also is directed perpendicuiar to and out 91" the page. The totai ﬁeld
at P is the vector sum of the two ﬁelds, and is of magnitude I B 3 (107 rim/Ammo A } (107 Tm/A)(2)(10 A) _ —5 . —5 __ —5
0.22m +WE27X10 T1419le0 Th3.6x1l} T. and is directed perpendicular to and out of the page. 20.51 From Table 20.1 on page 914 of the text, the magnitude of the magnetic ﬁeld at the center of a
circular current loop of radius r is no 271'] (10“7 Tm/A)(27r)(5.GA) _—
_ _ _. —————._M___ : , 1 a .
47:" 7‘ 10 X 10*2 m 3 1 X 0 T _ J this magnetic ﬁeld is directed along _ f— 1.602 x 1049 C](2.50 x 106 in/s)(3.1 x 10m5 T)sin 96° _ 13 2
c_ 911 X10_31 kg  1.4 x 10 m/s .
The acceleration is in the 3 sins direction as the force. Sin
the direction of \7‘ X B, i.e., in the direction of —i. Hence b) The magnitude of the magnetic ﬁeld a distance r from the bottom inﬁnite Wire is The direction of the ﬁel ' your right hand along the wire causing the ﬁeld
Since the magnetic ﬁeld ' ‘ ' ' ﬁeld of the bottom Wire at the position of the to " C) The force on a straight, current carrying Wire in a magnetic field that is constant "along its iength is
given by Equation 20.11 on page 905 of the text, .., .. Fzri‘xe. Fbomm = MB 2 {15A)(2.00 m )(m x 10—4 r) — USing the vector product right—hand rule to determine the direction of F on the bottom Wire, we see that it _'
is toward the top Wire. ' , "aw—r", W. FW—mmP—h—q ~ (l) The force on a straight, current carrying wire in a magnetic ﬁeld that is Constant along its length is
f : [Ex E. The top wire is perpendicnlar to the ﬁeld produced by the bottom wire, so the magnitude of the force on the top wire is Fm 2 MB 2 (10A)(2.00 m)(1.5 x 104 T) a 3.0 x 10""3 N. Using the vector product righthand rule to determine the direction of F on the top wire, we see that it is
toward the bottom wire. Note that the force of the top wire on the bottom wire and the force of the bottom wire on the top wire
form a Newton’s third law force pair. 20.65
a) The magnitude of the magnetic dipole moment is the loop in the direction of the current; the extended right hand thumb then points in the direction of i1. Doing this with the current loop shown in Figure P. 65 on page 948 of the text, we ﬁnd that [5 points into
the page, and thus, b) The torque on the current loop is
«r : ,2 x a : [—(0.450 Am2 )12] x [(4.00 mm (5.00 'r )1}; = MLSOINm‘i. (3) Since there is a nonzero torque on the loop, it will rotate. Place your right hand thumb along —i; your
ﬁngers then indicate the direction the loop will rotate about the yards. d) The potential energy of the 100p is PE = ﬁne 3‘ : 440.450 Am2)12}a{(4.00 T)i+ (5.00 mic] : 2.25 J. 7 This is not the minimum value of the potential energy, which is wh}r the 100p rotates.
e) Since the magnetic ﬁeld is uniform over the area of the loop, the magnetic ﬂux is '<E=§9A. The area vector is perpendicular to the loop, and in the same direction as Ii. Therefore, a = [(4.00 r)i+ (5.09 r)12}a[(0.150 m)2(—i2)] = 43.113 Trn. I:
:3
to
"H _ (10”? Tm/A)(2)(5.G A) _ *5
Btophwh50x10 T, (107 Tm/A )(2)(5.e A ) Bmiddle = 400 X104 m = 2.5 X 10 T, and (10? T.m/A)(2)(5.0 A) _5
6.00 X10“2 m 2 1'7 X 10 T' IBbottom : Therefor, the total magnetic ﬁeld created at P by the three currents is Biota! z Btop + Bmiddle + Bbottom = (5.0 x 10“5 TH: + (2.5 x 105 TJIE + (1.7 x 105 Tu; 2:. 3mm 2 {9.2 x 10*5 Tn}. ...
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 Spring '09
 HENRY
 Physics

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