CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS

Transcription

1 CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS PROBLEMS 5. SSM REASONING According to Equation 21.1, the agnitude of the agnetic force on a oving charge is F q 0 vb sinθ. Since the agnetic field points due north and the proton oves eastward, θ Furtherore, since the agnetic force on the oving proton balances its weight, we have g q 0 vb sinθ, where is the ass of the proton. This expression can be solved for the speed v. Solving for the speed v, we have v g q 0 B sinθ ( kg)(9.80 /s 2 ) ( C)( T) sin /s 8. REASONING There are two forces acting on the particle, an electric force and a agnetic force. We will deterine the agnitude and direction of each force. The net force is the vector su of these forces. a. The agnitude F electric of the electric force is equal to the product of the charge q and the agnitude E of the electric field (Equation 18.2), F electric qe ( C)123 ( N/C) N Since the charge is positive, the direction of the electric force is the sae as that of the electric field, along the x axis. The agnitude F agnetic of the agnetic force is given by Equation 21.1, F agnetic qvb sin θ ( C) ( /s)3.35 ( 10 5 T)sin N The direction of the agnetic force is found fro an application of Right-Hand Rule No. 1; the direction points along the +x axis. The net force F is F F electric + F agnetic N N N Negative, because the force points along the x axis b. Since the net force is positive, it points along the + x axis.

2 Chapter 21 Probles SSM WWW REASONING The direction in which the electrons are deflected can be deterined using Right-Hand Rule No. 1 and reversing the direction of the force (RHR-1 applies to positive charges, and electrons are negatively charged). Each electron experiences an acceleration a given by Newton s second law of otion, a F/, where F is the net force and is the ass of the electron. The only force acting on the electron is the agnetic force, F q 0 vb sin θ, so it is the net force. The speed v of the electron is related to its kinetic energy KE by the relation KE 1 2 v 2. Thus, we have enough inforation to find the acceleration. a. According to RHR-1, if you extend your right hand so that your fingers point along the direction of the agnetic field B and your thub points in the direction of the velocity v of a positive charge, your pal will face in the direction of the force F on the positive charge. For the electron in question, the fingers of the right hand should be oriented downward (direction of B) with the thub pointing to the east (direction of v). The pal of the right hand points due north (the direction of F on a positive charge). Since the electron is negatively charged, it will be deflected due south. b. The acceleration of an electron is given by Newton s second law, where the net force is the agnetic force. Thus, a F q 0 vb sinθ Since the kinetic energy is KE 1 2 v2, the speed of the electron is v acceleration of the electron is 2KE ( )/. Thus, the a q 0 vbsinθ q 0 2(KE) B sinθ ( ) ( C) J kg kg ( T)sin /s REASONING AND The charge can be found fro Equation 21.2 as q v Br ( )( /s) kg 0.75 T ( )( ) C Since e C, we see that the charge of the ionized heliu is +2e.

3 654 MAGNETIC FORCES AND MAGNETIC FIELDS 17. SSM WWW REASONING AND In one revolution, the particle oves once around the circuference of the circle. Therefore, it travels a distance of d 2π r, where r is the radius of the circle. Since the particle oves at constant speed, v d / t, and the tie required for one revolution is t d/ v. According to Equation 21.2, r v/ qb, so v qbr /. Thus, the tie required for the particle to coplete one revolution is t d v 2πr qbr / 2π B(q / ) 2π (0.72 T)( C/kg) s 23. REASONING AND The proton will iss the opposite plate if the distance between the plates is equal to the radius of the circular orbit of the proton. Therefore, B v qr ( kg) /s ( C)0.18 ( ) ( ) 0.13 T 32. REASONING The agnitude of the agnetic force exerted on a long straight wire is given by Equation as F ILB sin θ. The direction of the agnetic force is predicted by Right-Hand Rule No. 1. The net force on the triangular loop is the vector su of the forces on the three sides. a. The direction of the current in side AB is opposite to the direction of the agnetic field, so the angle θ between the is θ 180. The agnitude of the agnetic force is F AB ILB sin θ ILBsin 180 0N For the side BC, the angle is θ 55.0, and the length of the side is L agnetic force is 2.00 cos The F BC ILB sin θ ( 4.70 A) ( 3.49 ) ( 1.80 T)sin N An application of Right-Hand No. 1 shows that the agnetic force on side BC is directed perpendicularly out of the paper, toward the reader. For the side AC, the angle is θ 90.0, and the length of the side is L ( 2.00 )tan The agnetic force is F AC ILB sin θ ( 4.70 A) ( 2.86) ( 1.80 T)sin N An application of Right-Hand No. 1 shows that the agnetic force on side AC is directed perpendicularly into the paper, away fro the reader.

4 Chapter 21 Probles 655 b. The net force is the vector su of the forces on the three sides. Taking the positive direction as being out of the paper, the net force is F 0 N N + ( 24.2 N) 0N 36. REASONING AND When the agnetic field is turned on, a agnetic force that points downward (by Right-Hand Rule No. 1) is produced. This force ust be balanced by placing a ass on the pan such that Substituting in the nubers yields F NILB sin θ g so that NILB sin θ /g (125)(8.50 A)( )(0.200 T)(sin 90.0 ) 9.80 /s kg 41. SSM REASONING AND a. The agnetic oent of the coil is Magnetic oent NIA (50)(15 A) [ π(0.10 ) 2 ] 24 A 2 where we have used the fact that the area of the circular loop is A π r 2 π ( 0.10 ) 2. b. According to Equation 21.4, the torque is the product of the agnetic oent NIA and Bsinφ. However, the axiu torque occurs when φ 90.0, so we have τ ( Magnetic oent )( B sin 90.0 ) (24 A 2 )( 0.20 T) 4.8 N 44. REASONING AND The torque is given by τ NIAB sin φ. a. The axiu torque occurs when φ 90.0 (sin φ 1). In this case we want the torque to be 80.0% of the axiu value, so NIAB sinφ 0.800( NIAB sin90.0 ) so that φ sin 1 ( 0.800) REASONING AND Let the current in the left-hand wire be labeled I 1 and that in the right-hand wire. a. At point A: B 1 is up and B 2 is down, so we subtract the to get the net field. We have B 1 I /2pd 1 1 (8.0 A)/[2p(0.030 )]

5 656 MAGNETIC FORCES AND MAGNETIC FIELDS B 2 /2pd 2 (8.0 A)/[2p(0.150 )] So the net field at point A is B A B 1 B T b. At point B: B 1 and B 2 are both down so we add the two. We have So the net field at point B is B 1 (8.0 A)/[2p(0.060 )] B 2 (8.0 A)/[2p(0.060 )] B B B 1 + B T 60. REASONING AND The net force on the wire loop is a su of the forces on each segent of the loop. The forces on the two segents perpendicular to the long straight wire cancel each other out. The net force on the loop is therefore the su of the forces on the parallel segents (near and far). These are F n µ o I 1 L/2pd n µ o (12 A)(25 A)(0.50 )/[2p(0.11 )] N F f µ o I 1 L/2pd f µ o (12 A)(25 A)(0.50 )/[2p(0.26 )] N Note: F n is a force of attraction, while F f is a repulsive one. The agnitude of the net force is, therefore, F F n F f N N N 66. REASONING AND Apere's law in the for of Equation 21.8 indicates that ΣB l I. Since the agnetic field is everywhere perpendicular to the plane of the paper, it is everywhere perpendicular to the circular path and has no coponent B that is parallel to the circular path. Therefore, Apere's law reduces to ΣB l 0 I, so that the net current passing through the circular surface is zero.