Chapter 29 Solutions

Transcription

1 Chapter 29 Solutions 29.1 (a) up out of the page, since the charge is negative. (c) no deflection (d) into the page 29.2 At the equator, the Earth's agnetic field is horizontally north. Because an electron has negative charge, F q v B is opposite in direction to v B. Figures are drawn looking down. (a) (c) Down North East, so the force is directed West North North sin 0 0: Zero deflection West North Down, so the force is directed Up (d) Southeast North Up, so the force is Down (a) (c) (d) 29.3 F B q v B; F B ( j) e v i B Therefore, B B ( k) which indicates the negative z direction *29.4 (a) F B qvb sin θ ( C)( /s)( T) sin 37.0 F B N a F N kg /s F a ( kg)( /s 2 ) N qvb sin 90 B F qv N ( C)( /s) T The right-hand rule shows that B ust be in the y direction to yield a force in the +x direction when v is in the z direction by Harcourt, Inc. All rights reserved.

2 2 Chapter 29 Solutions *29.6 First find the speed of the electron: K 1 2 v2 e( V) U v 2e ( V ) 2400 J / C ( kg) C /s (a) F B, ax qvb ( C)( /s)(1.70 T) N F B, in 0 occurs when v is either parallel to or anti-parallel to B 29.7 Gravitational force: F g g ( kg)(9.80 /s 2 ) N down Electric force: Magnetic force: F e qe ( C)100 N/C down N up F B qv B C s E 50.0 N s 10 6 C N F B N up N down 29.8 We suppose the agnetic force is sall copared to gravity. Then its horizontal velocity coponent stays nearly constant. We call it v i. Fro v 2 y v 2 yi + 2a y ( y y i ), the vertical coponent at ipact is 2gh j. Then, F B qv B Qvi ( 2gh j) Bk QvB( j) Q 2gh Bi F B QvB vertical + Q 2gh B horizontal F B C(20.0 /s)( T) j C 2(9.80 /s 2 )(20.0 ) ( T) i F B ( N) vertical + ( N) horizontal 29.9 F B qvb sin θ so N ( C)( /s)(1.70 T) sin θ sin θ and θ sin -1 (0.754) 48.9 or 131

3 Chapter 29 Solutions qe ( C) ( 20.0 N / C)k ( N)k ΣF qe + qv B a ( N)k C ( /s i) B ( )( /s 2 )k ( N)k ( C /s)i B ( N)k ( C /s)i B ( N)k The agnetic field ay have any x-coponent. B z 0 and B y 2.62 T F B qv B i j k v B ( 12 2)i + ( 1 + 6)j + ( 4 + 4)k 10i + 7 j + 8k v B T /s F B q v B ( C) ( 14.6 T s) N F B qv B i j k F B ( C) [( 0 0)i + ( 0 0)j + ( 0 ( 1.40 T) ( s) ) k] ( k)n F B ILB sin θ with F B F g g g ILB sin θ so L g IB sin θ I 2.00 A and L 100 c / ( g / c) 1000 g / kg kg / Thus ( )(9.80) (2.00)B sin 90.0 B Tesla with the direction given by right-hand rule: eastward 2000 by Harcourt, Inc. All rights reserved.

4 4 Chapter 29 Solutions Goal Solution A wire having a ass per unit length of g/c carries a 2.00-A current horizontally to the south. What are the direction and agnitude of the iniu agnetic field needed to lift this wire vertically upward? G : Since I 2.00 A south, B ust be to the east to ake F upward according to the right-hand rule for currents in a agnetic field. The agnitude of B should be significantly greater than the earth s agnetic field (~ 50 µt), since we do not typically see wires levitating when current flows through the. O : The force on a current-carrying wire in a agnetic field is F B I1 B, fro which we can find B. A : With I to the south and B to the east, the force on the wire is siply F B I lbsin90, which ust oppose the weight of the wire, g. So, B F B Il g Il g I l 9.80 / s A g 10 2 c / c 10 3 g / kg T L : The required agnetic field is about 5000 ties stronger than the earth s agnetic field. Thus it was reasonable to ignore the earth s agnetic field in this proble. In other situations the earth s field can have a significant effect F B I L B (2.40 A)(0.750 )i (1.60 T)k ( 2.88 j) N (a) F B ILB sin θ (5.00 A)(2.80 )(0.390 T) sin N F B (5.00 A)(2.80 )(0.390 T) sin N (c) F B (5.00 A)(2.80 )(0.390 T) sin N F B L g L I L B L I g BL ( kg/)(9.80 /s2 ) 3.60 T A The direction of I in the bar is to the right.

5 Chapter 29 Solutions The agnetic and gravitational forces ust balance. Therefore, it is necessary to have F B BIL g, or I (g/bl) (λ g/b) [λ is the ass per unit length of the wire]. Thus, I ( kg/)(9.80 /s 2 ) ( T) 196 A (if B 50.0 µt) The required direction of the current is eastward, since East North Up For each segent, I 5.00 A and B N / A j Segent L F B I( L B) ab bc cd da j k i j i k 0 (40.0 N)( i) (40.0 N)( k) (40.0 N)(k + i) The rod feels force F B I( d B) Id( k) B( j) IdB() i The work-energy theore is ( K trans + K rot ) i + E ( K trans + K rot ) f Fscosθ 1 2 v Iω 2 IdBLcos v v R2 R 2 and IdBL 3 4 v2 v 4IdBL 3 4(48.0 A)(0.120 )(0.240 T)(0.450 ) 1.07 /s 3(0.720 kg) The rod feels force F B I( d B) Id( k) B( j) IdB() i The work-energy theore is ( K trans + K rot ) i + E ( K trans + K rot ) f Fscosθ 1 2 v Iω 2 IdBLcos v v R2 R 2 and v 4IdBL by Harcourt, Inc. All rights reserved.

6 6 Chapter 29 Solutions The agnetic force on each bit of ring is I ds B I ds B radially inward and upward, at angle θ above the radial line. The radially inward coponents tend to squeeze the ring but all cancel out as forces. The upward coponents I ds B sin θ all add to I 2π rb sin θ up. *29.22 Take the x-axis east, the y-axis up, and the z-axis south. The field is B ( 52.0 µt)cos 60.0 ( k)+ ( 52.0 µt)sin 60.0 ( j) The current then has equivalent length: L 1.40 ( k) () j F B I L B ( A) ( 0.850j k) ( 45.0j 26.0k)10 6 T F B N( 22.1i 63.0i) N( i) 2.98 µn west (a) 2π r 2.00 so r µ IA ( A) [ π(0.318) 2 2 ] 5.41 A 2 τ µ B so τ ( A 2 )(0.800 T) 4.33 N *29.24 τ µb sin θ so N µ(0.250) sin 90.0 µ A A τ NBAI sin θ 1.20 A τ 100( T) τ 9.98 N sin 60 Note that θ is the angle between the agnetic oent and the B field. The loop will rotate so as to align the agnetic oent with the B field. Looking down along the y-axis, the loop will rotate in a clockwise direction. (a)

7 Chapter 29 Solutions (a) Let θ represent the unknown angle; L, the total length of the wire; and d, the length of one side of the square coil. Then, use the right-hand rule to find µ NAI L 4d d2 I at angle θ with the horizontal. At equilibriu, Στ (µ B) (r g) 0 ILBd 4 sin(90.0 θ) gd 2 sinθ 0 and gd 2 sinθ θ tan 1 ILB 2g (3.40 A)(4.00 )( T) tan 1 2(0.100 kg)(9.80 / s 2 ) 3.97 ILBd 4 cosθ τ ILBd 4 cosθ 1 (3.40 A)(4.00 )( T)(0.100 ) cos N Fro τ µ B IA B, the agnitude of the torque is IAB sin 90.0 (a) Each side of the triangle is 40.0 c/3. Its altitude is c 11.5 c and its area is A 1 2 (11.5 c)(13.3 c) Then τ (20.0 A)( )(0.520 N s/c ) 80.1 N Each side of the square is 10.0 c and its area is 100 c τ (20.0 A)( )(0.520 T) N (c) r /2π A π r τ (20.0 A)( )(0.520) N (d) The circular loop experiences the largest torque. *29.28 Choose U 0 when the dipole oent is at θ 90.0 to the field. The field exerts torque of agnitude µbsinθ on the dipole, tending to turn the dipole oent in the direction of decreasing θ. Its energy is given by U 0 θ 90.0 µbsinθ dθ θ µb( cosθ) 90.0 µbcosθ + 0 or U µ B 2000 by Harcourt, Inc. All rights reserved.

8 8 Chapter 29 Solutions *29.29 (a) The field exerts torque on the needle tending to align it with the field, so the iniu energy orientation of the needle is: pointing north at 48.0 below the horizontal where its energy is U in µbcos ( 10 3 A 2 )( T) J It has axiu energy when pointing in the opposite direction, south at 48.0 above the horizontal where its energy is U ax µbcos ( 10 3 A 2 )( T) J U in + W U ax : W U ax U in J ( J) 1.07 µj (a) τ µ B, so τ µ B µb sin θ NIAB sin θ τ ax NIABsin A [ ] T π( ) µn U µ B, so µb U +µb Since µb ( NIA)B A [ ] T π( ) µj, the range of the potential energy is: 118 µj U +118 µj (a) B T; v /s Direction is given by the right-hand-rule: southward F B qvb sin θ F B ( C)( /s)( T) sin N F v2 r so r v2 F ( kg)( /s) N 1.29 k (a) 1 2 v2 q( V) 1 2 ( kg) v 2 ( C)(833 V) v 91.3 k/s The agnetic force provides the centripetal force: qvb sin θ v2 r

9 Chapter 29 Solutions 9 r v qb sin 90.0 ( kg)( /s) ( C)(0.920 N s/c ) 1.98 c 2000 by Harcourt, Inc. All rights reserved.

10 10 Chapter 29 Solutions For each electron, q vb sin 90.0 v2 r and v ebr The electrons have no internal structure to absorb energy, so the collision ust be perfectly elastic: K 2 1 v 1i v 1f v 2 2 f K 2 1 e2 B 2 2 R e2 B 2 2 R 2 2 e2 B 2 2 R R 2 K e( C)( N s/c ) 2 2( kg) [( ) 2 + ( ) 2 ] 115 kev We begin with qvb v2 R qrb, so v The tie to coplete one revolution is T 2πR v 2πR qrb 2π qb Solving for B, B 2π qt T q( V) 2 1 v2 or v 2q ( V ) Also, qvb v2 r so r v qb qb 2q( V) 2 ( V ) qb 2 Therefore, r 2 p 2 p( V) eb 2 r 2 d 2 d ( V) q d B 2 V 22 p eb p V eb 2 2r p 2 and r 2 α 2 α ( V) q α B 2 V 24 p ( 2e)B p V eb 2 2r p 2 The conclusion is: r α r d 2 r p

11 Chapter 29 Solutions 11 Goal Solution A proton (charge +e, ass p), a deuteron (charge +e, ass 2 p ), and an alpha particle, (charge + 2e, ass 4 p ) are accelerated through a coon potential difference V. The particles enter a unifor agnetic field B with a velocity in a direction perpendicular to B. The proton oves in a circular path of radius r p. Deterine the values of the radii of the circular orbits for the deuteron r d and the alpha particle r α in ters of r p. G : In general, particles with greater speed, ore ass, and less charge will have larger radii as they ove in a circular path due to a constant agnetic force. Since the effects of ass and charge have opposite influences on the path radius, it is soewhat difficult to predict which particle will have the larger radius. However, since the ass and charge ratios of the three particles are all siilar in agnitude within a factor of four, we should expect that the radii also fall within a siilar range. O : The radius of each particle s path can be found by applying Newton s second law, where the force causing the centripetal acceleration is the agnetic force: F qv B. The speed of the particles can be found fro the kinetic energy resulting fro the change in electric potential given. A : An electric field changes the speed of each particle according to ( K + U) i ( K + U) f. Therefore, assuing that the particles start fro rest, we can write q V 1 2 v2. The agnetic field changes their direction as described by ΣF a: thus For the protons, For the deuterons, For the alpha particles, qvbsin 90 v2 r r v qb 2q V qb r p 1 B r d 1 B r α 1 B 1 B 2 p V e 2(2 p ) V 2r p e 2(4 p ) V 2r p 2e 2 V q L : Soewhat surprisingly, the radii of the deuterons and alpha particles are the sae and are only 41% greater than for the protons (a) We begin with qvb v2 R, or qrb v. But, L vr qr2 B. Therefore, R L qb J s ( C) T c Thus, v L R J s kg s 2000 by Harcourt, Inc. All rights reserved.

12 12 Chapter 29 Solutions ω qb ( C)(5.20 T) kg rad/s v2 q( V) so v 2q ( V ) r v qb so r 2q( V)/ qb r 2 q 2( V) B 2 and r 2 q 2( V) B 2 qb2 r 2 2( V) and ( ) ( q )B 2 r 2( V) 2 so 2 q q r r 2 2e e 2R R E 1 2 v2 e( V) and evbsin90 v 2 R B v er er 2e( V) 1 R 2( V) e 1 B ( kg)( V) T C r v qb so rqb v ( C) 1.80 T s u kg kg 2.99 u The particle is singly ionized: either a tritiu ion, 1 3 H +, or a heliu ion, 2 3 He F B F e so qvb qe where v 2K /. K is kinetic energy of the electrons E vb 2K B /2 ( ) 244 kv/

13 Chapter 29 Solutions K 1 2 v2 q( V) so v 2q ( V ) F B qv B v2 r r v qb q 2q( V)/ B 1 B 2( V) q (a) r 238 2( ) c r c r 238 r The ratios of the orbit radius for different ions are independent of V and B In the velocity selector: v E B 2500 V T s In the deflection chaber: r v qb ( s) ( T) kg C K 1 2 v2 : ( ev) ( J / ev) ( kg)v 2 v /s r v qb ( kg)( /s) ( C)(5.20 T) (a) F B qvb v2 R ω v R qbr R qb ( C)(0.450 T) kg rad/s v qbr ( C)(0.450 T)(1.20 ) kg /s F B qvb v2 r B v qr kg /s ( C)(1000 ) 3.00 T 2000 by Harcourt, Inc. All rights reserved.

14 14 Chapter 29 Solutions θ tan and R 1.00 c sin c Ignoring relativistic correction, the kinetic energy of the electrons is 1 2 v2 q( V) so v 2q ( V ) /s Fro the centripetal force v2 R field qvb, we find the agnetic B v q R ( kg)( /s) ( C)( T ) (a) R H 1 nq so n 1 qr H 1 ( C) C V H IB nqt ( C) ( ) ( V) B nqt ( V H) I 20.0 A 1.79 T nq t V H IB ( V)( ) (21.0 A)(1.80 T) /C Since V H IB nqt, and given that I 50.0 A, B 1.30 T, and t 0.330, the nuber of charge carriers per unit volue is n IB e( V H )t The nuber density of atos we copute fro the density: n g c 3 1 ole 63.5 g atos ole 106 c ato / 3 So the nuber of conduction electrons per ato is n n

15 29.51 B nqt ( V H) I B T 43.2 µt ( C) ( ) ( V) 8.00 A Chapter 29 Solutions 15 Goal Solution In an experient designed to easure the Earth's agnetic field using the Hall effect, a copper bar c thick is positioned along an east-west direction. If a current of 8.00 A in the conductor results in a Hall voltage of 5.10 pv, what is the agnitude of the Earth's agnetic field? (Assue that n electrons/ 3 and that the plane of the bar is rotated to be perpendicular to the direction of B.) G : The Earth s agnetic field is about 50 µt (see Table 29.1), so we should expect a result of that order of agnitude. O : The agnetic field can be found fro the Hall effect voltage: V H IB nqt or B nqt V H I A : Fro the Hall voltage, ( B e - 3 )( Ce - ) A V T 43.2 µt L : The calculated agnetic field is slightly less than we expected but is reasonable considering that the Earth s local agnetic field varies in both agnitude and direction (a) V H IB nqt so nqt I B T V H V 1.14 T 105 V Then, the unknown field is B nqt I V H B ( TV) ( V) T 37.7 T nqt I T V so n T I V qt n T A V C ( ) by Harcourt, Inc. All rights reserved.

16 16 Chapter 29 Solutions q vb sin 90 v2 r ω v r eb θ t (a) The tie it takes the electron to coplete π radians is t θ ω θ eb (π rad)( kg) ( C)(0.100 N s/c ) s Since v q Br, K e 1 2 v2 q2 B 2 r 2 2 e( C)(0.100 N s / C) 2 ( ) 2 2( kg) 351 kev F y 0: +n g 0 F x 0: µ k n + IBd sin B µ kg Id 0.100(0.200 kg)(9.80 /s2 ) (10.0 A)(0.500 ) 39.2 T (a) The electric current experiences a agnetic force. I(h B) in the direction of L. The sodiu, consisting of ions and electrons, flows along the pipe transporting no net charge. But inside the section of length L, electrons drift upward to constitute downward electric current J (area) JLw. The current then feels a agnetic force I h B JLwhB sin 90 This force along the pipe axis will ake the fluid ove, exerting pressure F area JLwhB hw JLB The agnetic force on each proton, F B qv B qvb sin 90 downward perpendicular to velocity, supplies centripetal force, guiding it into a circular path of radius r, with qvb v2 r and r v qb We copute this radius by first finding the proton's speed: K 1 2 v2 ( J / ev) v 2K ev kg /s

17 Chapter 29 Solutions 17 Now, r v qb ( kg)( /s)(c ) ( C)( N s) Fro the figure, observe that sin 1.00 r (a) α 8.90 The agnitude of the proton oentu stays constant, and its final y coponent is ( kg)( /s) sin(8.90 ) kg /s *29.57 (a) If B B x i + B y j + B z k, F B qv B ev ( i i) ( B x i + B y j + B z k) 0 + ev i B y k ev i B z j Since the force actually experienced is F B F i j, observe that B x could have any value, B y 0, and B z F i ev i If v v i i, then F B qv B e( v i i) ( B x i + 0j F i ev i k) F i j (c) If q e and v v i i, then F B qv B ev ( i i) ( B x i + 0j F i ev i k) F i j Reversing either the velocity or the sign of the charge reverses the force A key to solving this proble is that reducing the noral force will reduce the friction force: F B BIL or B F B IL When the wire is just able to ove, ΣF y n + F B cosθ g 0 so and n g F B cosθ f µ ( g F B cosθ) Also, ΣF x F B sinθ f 0 so F B sinθ f : We iniize B by iniizing F B : F B sinθ µ ( g F B cosθ) and F B df B dθ µg cosθ µ sinθ ( sinθ + µ cosθ) µg sinθ + µ cosθ 2 0 µ sinθ cosθ Thus, θ tan 1 1 tan 1 ( 5.00) 78.7 for the sallest field, and µ ( L) B F B IL µg I sinθ + µ cosθ 9.80 s2 B in A kg sin ( 0.200)cos T B in T pointing north at an angle of 78.7 below the horizontal 2000 by Harcourt, Inc. All rights reserved.

18 18 Chapter 29 Solutions (a) The net force is the Lorentz force given by F qe + qv B q( E + v B) 4i 1j 2k F [+ ( 2i + 3j 1k) ( 2i + 4j + 1k) ]N Carrying out the indicated operations, we find: F ( 3.52i 1.60 j) N θ cos 1 F x F 3.52 cos ( 1.60) r v qb ( )( ) ( )( ) k No, the proton will not hit the Earth Let x 1 be the elongation due to the weight of the wire and let x 2 be the additional elongation of the springs when the agnetic field is turned on. Then F agnetic 2k x 2 where k is the force constant of the spring and can be deterined fro k g/2 x 1. (The factor 2 is included in the two previous equations since there are 2 springs in parallel.) Cobining these two equations, we find F agnetic 2 g x 2 g x 2 ; but F B I L B ILB 2 x 1 x 1 Therefore, where I 24.0 V 12.0 Ω 2.00 A, B g x 2 IL x 1 (0.0100)(9.80)( ) (2.00)(0.0500)( ) T *29.62 Suppose the input power is 120 W ( 120 V)I : I ~1 A 10 0 A Suppose ω 2000 rev 1 in 2π rad in 60 s 1 rev ~200 rad s and the output power is 20 W τω τ 200 rad s τ ~10 1 N Suppose the area is about ( 3 c) ( 4 c), or A~ Fro Table 29.1, suppose that the field is B~10 1 T Then, the nuber of turns in the coil ay be found fro τ NIAB: 0.1 N ~N 1 C s 10 1 N s C giving N ~10 3

19 Chapter 29 Solutions Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibriu: ΣF x T sinθ ILB sin or T sinθ ILB ΣF y T cosθ g 0, or T cosθ g Therefore, tanθ ILB g IB ( L)g 5.00 A ( kg ) 9.80 s2 B or tan( 45.0 ) 19.6 T ( B L)g tanθ I Call the length of the rod L and the tension in each wire alone T 2. Then, at equilibriu: ΣF x T sinθ ILBsin or T sinθ ILB ΣF y T cosθ g 0, or T cosθ g tanθ ILB g IB ( L)g or ( B L)g tanθ I µg I tanθ ΣF a or qvb sin 90.0 v2 r the angular frequency for each ion is v r ω qb 2π f and f f 12 f 14 qb 1 1 ( C)(2.40 T) 1 2π 12 2π( kg / u) 12.0 u u 14 f f 12 f s khz Let v x and v be the coponents of the velocity of the positron parallel to and perpendicular to the direction of the agnetic field. (a) The pitch of trajectory is the distance oved along x by the positron during each period, T (see Equation 29.15). p v x T (vcos 85.0 ) 2π Bq p ( )(cos 85.0 )(2π)( ) (0.150)( ) Fro Equation 29.13, r v Bq v sin 85.0 Bq r ( )( )(sin 85.0 ) (0.150)( ) by Harcourt, Inc. All rights reserved.

20 20 Chapter 29 Solutions τ IAB where the effective current due to the orbiting electrons is I q q t T and the period of the otion is T 2πR v The electron's speed in its orbit is found by requiring k e q2 R 2 v2 R or v q k e R Substituting this expression for v into the equation for T, we find T 2π R3 q 2 k e T 2π ( )( ) 3 ( ) 2 ( ) s Therefore, τ q T AB π( ) 2 (0.400) N Goal Solution Consider an electron orbiting a proton and aintained in a fixed circular path of radius R by the Coulob force. Treating the orbiting charge as a current loop, calculate the resulting torque when the syste is in a agnetic field of T directed perpendicular to the agnetic oent of the electron. G : Since the ass of the electron is very sall (~10-30 kg), we should expect that the torque on the orbiting charge will be very sall as well, perhaps ~10-30 N. O : The torque on a current loop that is perpendicular to a agnetic field can be found fro τ IAB sinθ. The agnetic field is given, θ 90, the area of the loop can be found fro the radius of the circular path, and the current can be found fro the centripetal acceleration that results fro the Coulob force that attracts the electron to proton. A : The area of the loop is A π r 2 π ( ) If v is the speed of the electron, then the period of its circular otion will be T 2πR v, and the effective current due to the orbiting electron is I Q / t et. Applying Newton s second law with the Coulob force acting as the central force gives ΣF k e q2 R 2 v2 R so that v q k e R and T 2π R3 q 2 k e T 2π ( kg)( ) 3 ( C) N 2 C s The torque is τ q T AB: τ C s (π)( ) 2 (0.400 T) N L : The torque is certainly sall, but a illion ties larger than we guessed. This torque will cause the ato to precess with a frequency proportional to the applied agnetic field. A siilar process on the nuclear, rather than the atoic, level leads to nuclear agnetic resonance (NMR), which is used for agnetic resonance iaging (MRI) scans eployed for edical diagnostic testing (see Section 44.2).

21 Chapter 29 Solutions Use the equation for cyclotron frequency ω qb or qb ω qb 2π f ( C)( T) (2π)(5.00 rev / s) kg J (a) K 2 1 v MeV ev K J v ( J) kg s F B qvb v2 R so R v qb kg ev ( s) ( C) ( 1.00 T) x v θ x x x x x x Bx in x 1.00 x T x x x x x x 45 x x x x x 45 x R x x x x x x x x x x x x x x 45.0 x x x x x Then, fro the diagra, x 2R sin ( )sin Fro the diagra, observe that θ' (a) See graph to the right. The Hall voltage is directly proportional to the agnetic field. A least-square fit to the data gives the equation of the best fitting line as: V H ( VT)B Coparing the equation of the line which fits the data best to I V H nqt B observe that: V H (µv) I nqt VT, or t I nq VT Then, if I A, q C, and n , the thickness of the saple is A t ( ) C B (T) ( VT) by Harcourt, Inc. All rights reserved.

22 22 Chapter 29 Solutions *29.71 (a) The agnetic force acting on ions in the blood strea will deflect positive charges toward point A and negative charges toward point B. This separation of charges produces an electric field directed fro A toward B. At equilibriu, the electric force caused by this field ust balance the agnetic force, so or qvb qe q V d v V Bd V T /s N o. Negative ions oving in the direction of v would be deflected toward point B, giving A a higher potential than B. Positive ions oving in the direction of v would be deflected toward A, again giving A a higher potential than B. Therefore, the sign of the potential difference does not depend on whether the ions in the blood are positively or negatively charged. *29.72 When in the field, the particles follow a circular path according to qvb v 2 r, so the radius of the path is: r v / qb (a) (c) When r h v qbh, that is, when v qb, the particle will cross the band of field. It will ove in a full seicircle of radius h, leaving the field at ( 2h,0,0) with velocity v f vj. When v < qbh v, the particle will ove in a saller seicircle of radius r < h. It will qb leave the field at 2r,0,0 with velocity v f v j. When v > qbh v, the particle oves in a circular arc of radius r > h, centered at ( r,0,0 qb ). The arc subtends an angle given by θ sin 1 ( hr). It will leave the field at the point with coordinates r( 1 cosθ), h,0 [ ] with velocity v f vsinθ i + vcosθ j.