{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

obj-12

# obj-12 - 27.1 27.2.W,W,~c.mxmewsmmﬁm,“Kw.m-»w IDENTIFY...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 27.1., 27.2. .._.W,W,~c.,.mxmewsmmﬁm,“Kw.m-»w IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F. Use the cross products of unit vectors from Section 1.10. EXECUTE: i=(+4.l9><104 m/s)i°+(—3.85x104 iii/s); (a) B =(1.40 T)? F =qi7><é =(—1.24><1(r8 C)(1.40 T)[(4.19x104 m/s)i><§—(3.85><104 m/s)j‘><§] "WWW ixi=0jxf=4€ F‘ =(—1.24><10-8 c)(1.40 T)(—3.85x10“ iii/s)(—IE)=(—6.68><10*1 N)I€ EVALUATE: The directions of v‘ and I} are shown in Figure 27.1a. The right-hand rule gives that ﬁx]? is directed out of the paper (+z-direction). The charge is negative so F is opposite to 17 X3; Figure 27.1a F is in the —-z»— direction. This agrees with the direction calculated with unit vectors. (b) EXECUTE: E = (1.40 T)I€ 17* = qrxé = (—1.24><1o-8 c)(1.40 T)[(+4.19X10“ iii/s)? x12 —(3.85><10“ m/s) 2x12] A fxlfr:—j‘,j‘xl€=i I7 =(—7.27x10*‘ N)(—f)+(6.68><10‘4 N)€=[(6.68x10“‘ N)i+(7.27x10'4 N) j] EVALUATE: The directions of I7 and i? are shown in Figure 27.1b. The direction of I7 is opposite to 17 X1} since x i q is negative. The direction of F computed / \v from the right-hand rule agrees qualitatively with the direction calculated with unit vectors. 3” x1? (by right-hand rule) Figure 27.1b IDENTIFY: The net force must be zero, so the magnetic and gravity forces must be equal in magnitude and opposite in direction. SET UP: The gravity force is downward so the force from the magnetic ﬁeld must be upward. The charge’s velocity and the forces are shown in Figure 27.2. Since the charge is negative, the magnetic force is opposite to the ri t-hand rule direction. The minimum magnetic ﬁeld is when the ﬁeld is perpendicular to 17 . The force is also perpendicular to E , so I} is either eastward or westward. EXECUTE: If E is eastward, the right-hand rule direction is into the page and I78 is out of the page, as required; —3 2 Therefore, 3 is eastward. mg = lqlvBsin¢. ¢ = 90° and B = ﬂ = W =1_91 T , vqu (4.00x10 m/s)(2.50x10 C) EVALUATE: The magnetic field could also have a component along the north~south direction, that would not contribute to the force, but then the field wouldn’t have minimum magnitude. N W «—{—» E i? S 17 ® (9 F3; Figure 27.2 "N 27.4. IDENTIFY: Apply Newton’s second law, with the force being the magnetic force. SET UP: fxi: ~13 _. g tiXB EXECUTE: F = mi 2 qt? X3 gives a = and m (1.22x10'8 C)(3.0><104 m/s)(1.63 T)(j'xi‘) _3 = —(0.330 m/s2)12. 1.81x10 kg [i : EVALUATE: The acceleration is in the ~z-direction and is perpendicular to both i7 and B . 27.5. IDENTIFY: Apply F = IqlvBsin¢ and solve for v. SET UP: An electron has q : —1.60X10‘l9 C. _ F _ 4.60><10‘l5 N |qlBsin¢ (1.6x10~19 C)(3.5><10‘3 T)sin60° EVALUATE: Only the component Bsin¢ of the magnetic field perpendicular to the velocity contributes to the force. 27.6. IDENTIFY: Apply Newton’s second law and F =lqlvBsin¢. EXECUTE: v = 9.49><106 m/s SET UP: ¢ is the angle between the direction of i” and the direction of B . EXECUTE: (a) The smallest possible acceleration is zero, when the motion is parallel. to the magnetic ﬁeld. The greatest acceleration is when the velocity and magnetic ﬁeld are at right angles: -19 6 ~2 a =M=www=mmwa m (9.11X10 kg) (b) If a=%(3.25><1016 m/s2)=513§§i"¥'3, then sin¢=0.25 and ¢=14.5°. m EVALUATE: The force and acceleration decrease as the angle ¢ approaches zero. ‘ 27.15. (a) IDENTIFY: Apply Eq.(27.2) to relate the magnetic force F" to the directions of 17 and 3. The electron has negative charge so F is opposite to the direction of .17 X1}. For motion in an arc of a circle the acceleration is toward the center of the are so I?" must be in this direction. a = v2 IR. SET UP: {1’0 x “é V v0 0 As the electron moves in the semicircle, its velocity is tangent to the circular path. R F The direction of 170 XB at a point along A B B the path is shown in Figure 27.15. (-——---—> 0.100 m Figure 27.15 EXECUTE: For circular motion the acceleration of the electron rim is directed in toward the center of the circle. Thus the force 133 exerted by the magnetic ﬁeld, since it is the only force on the electron, must be radially inward. _. .. Since q is negative, F B is opposite to the direction given by the right—hand rule for 170 x5. Thus B is directed into the page. Apply Newton's 2nd law to calculate the magnitude of 3: 217 = mii gives 2F,“ = ma FB = m(v2/R) FB = lqlvBsin¢ = lqlvB, so IqlvB = m(v2/R) = ﬂ ___ (9.109x10‘3' kg)(1.41x10‘5 m/s) |qu (1.602x10'” C)(0.050 m) (b) IDENTIFY and SET UP: The speed of the electron as it moves along the path is constant. (F's changes the direction of 17 but not its magnitude.) The time is given by the distance divided by v0. ' EXECUTE: The distance along the semicircular path is ﬂ'R, so I = :0- = = 1.11x10'7 s B =1.6()><10~4 T EVALUATE: The magnetic field required increases when v increases or R decreases and also depends on the mass to charge ratio of the particle. 27.22.. 27.28. 2 V IDENTIFY: For motion in an arc of a circle, a :76 and the net force is radially inward, toward the center of the circle. SET UP: The direction of the force is shown in Figure 27.22. The mass of a proton is 1.67X10‘27 kg . EXECUTE: (a) F is opposite to the right—hand rule direction, so the charge is negative. F = mi gives . 2 BR 1. "9 . . ‘q'vBSID¢=mv_. ¢=90°and 1}:qu m/S ‘ R m 12(1.67X10 kg) (h) FB = [q|vBsin¢ = 3(1.60><10“9 C)(2.84><1o6 m/s)(0.250 T)sin90° = 3.41x10‘” N. w = mg =12(1.67 ><10”27 kg)(9.80 111/32)=1.96X10"25 N . The magnetic force is much larger than the weight of the particle, so it is a very good approximation to neglect gravity. EVALUATE: (c) The magnetic force is always perpendicular to the path and does no work. The particles move with constant speed. 0 0 I? O O O I O O O O 0 Figure 27.22 IDENTIFY: For no deﬂection the magnetic and electric forces must be equal in magnitude and opposite in direction. SET UP: v = E/ B for no deﬂection. With only the magnetic force, lqlvB = mvz / R EXECUTE: (a) v = E/B =(1.56x10“V/m)/(4.62x10“3 T) = 3.38X106m/s. (b) The directions of the three vectors 17, I73 and I} are sketched in Figure 27.28. ~31 6 (c) R: mv : (9.11x10 _19kg)(3.38><10 3m/s) =4‘17X104 m. |q|B (1.60X10 C)(4.62x10' T) —3 T:27m_271R=27r(4.17><10 m)=7‘74xlo_gs' [ng ' v (3.38x106m/s) EVALUATE: For the ﬁeld directions shown in Figure 27.28, the electric force is toward the top of the page and the magnetic force is toward the bottom of the page. Figure 27.28 27.33. 27.34. 27.40. IDENTITY: The magnetic force is F = [[3 sin a). For the wire to be completely supported by the field requires that F = mg and that 15' and W are in opposite directions. SET UP: The magnetic force is maximum when ¢ 2 90°. The gravity force is downward. (a) 113 — mg I ~35: —— (0'150 ngg-go m/SZ) V [B (2.00 m)(0.55><10‘4 T) heating due to the resistance of the wire would be severe; such a current isn’t feasible. (b) The magnetic force must be upward. The directions of I, B and F‘ are shown in Figure 27.33, where we have assumed that B is south to north. To produce an upward magnetic force, the current must be to the east. The wire must be horizontal and perpendicular to the earth’s magnetic field. EVALUATE: The magnetic force is perpendicular to both the direction of I and the direction of E. N EXECUTE: 1.34><104 A. This is a very large current and ohmic B S Figure 27.33 IDENTIFY: Apply F = [[3 sin ¢. SET UP: 1 = 0.0500 In is the length of wire in the magnetic field. Since the wire is perpendicular to E, ¢ = 90°. EXECUTE: F = 118 = (10.8 A)(0.0500 m)(0.550 T) = 0.297 N. EVALUATE: The force per unit length of wire is proportional to both B and I. IDEIyTIFY: The magnetic force F), must be upward and equal to mg. The direction of 17'” is detemiined by the direction of I in the circuit. ‘ . . V SET UP: FB = IlBsm¢ , With ¢ = 90° . I = E , where Vis the battery voltage. EXECUTE: (a) The forces are shown in Figure 27.40. The current I in the bar must be to the right to produce F3 upward. To produce current in this direction, point a must be the positive terminal of the battery. (b) FB = mg . [13 2mg . m = g5 : E13 = (175 V)(0.600 m)(l.50 T) g Rg (5.00 £2)(9.80 m/sz) EVALUATE: If the battery had opposite polarity, with point a as the negative terminal, then the current would be clockwise and the magnetic force would be downward. :3.21 kg . F3 369 mg Figure 27.40 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

obj-12 - 27.1 27.2.W,W,~c.mxmewsmmﬁm,“Kw.m-»w IDENTIFY...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online