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27.3. IDENTIFY: The force F on the particle is in the direction of the deflection of the particle. Apply the right-hand rule to the directions of v and B . See if your thumb is in the direction of F , or opposite to that direction. Use F q vB sin with 90° to calculate F. SET UP: The directions of v , B and F are shown in Figure 27.3. EXECUTE: (a) When you apply the right-hand rule to v and B , your thumb points east. F is in this direction, so the charge is positive. (b) F q vB sin (8.50 106 C)(4.75 103 m/s)(1.25 T)sin90° 0.0505 N EVALUATE: If the particle had negative charge and v and B are unchanged, the particle would be deflected toward the west. Figure 27.3 27.5. IDENTIFY: Apply F q vB sin and solve for v. SET UP: An electron has q 1.60 1019 C . EXECUTE: v F 4.60 1015 N 9.49 106 m s 19 q B sin (1.6 10 C)(3.5 103 T)sin 60 EVALUATE: Only the component B sin of the magnetic field perpendicular to the velocity contributes to the force. 27.7. IDENTIFY: Apply F = qv B . SET UP: v = vy ˆj , with vy 3.80 103 m s . Fx 7.60 103 N, Fy 0, and Fz 5.20 103 N . EXECUTE: (a) Fx q(v y Bz vz By ) qv y Bz . Bz Fx qvy (7.60 103 N) ([7.80 106 C)( 3.80 103 m s )] 0.256 T Fy q(vz Bx vx Bz ) 0, which is consistent with F as given in the problem. There is no force component along the direction of the velocity. Fz q(vx By v y Bx ) qv y Bx . Bx Fz qv y 0.175 T . (b) By is not determined. No force due to this component of B along v ; measurement of the force tells us nothing about B y . (c) B F Bx Fx By Fy Bz Fz (0.175 T)(+7.60 103 N) (0.256 T)(5.20 103 N) B F 0 . B and F are perpendicular (angle is 90) . EVALUATE: The force is perpendicular to both v and B , so v F is also zero. 27.9. IDENTIFY: Apply F qv B to the force on the proton and to the force on the electron. Solve for the components of B . SET UP: F is perpendicular to both v and B . Since the force on the proton is in the +y-direction, By 0 and B = Bx iˆ Bz kˆ . For the proton, v = (1.50 km/s)iˆ . EXECUTE: (a) For the proton, F = q(1.50 103 m/s)iˆ (Bx iˆ Bz kˆ) q(1.50 103 m/s)Bz ( ˆj). 2.25 1016 N 0.938 T . The force on the proton is (1.60 1019 C)(1.50 103 m/s) independent of Bx . For the electron, v = (4.75 km/s)( kˆ ) . F = (2.251016 N) ˆj , so Bz F qv B (e)(4.75 103 m/s)(kˆ) ( Bxiˆ Bz kˆ ) = e(4.75 103 m/s) Bx ˆj . The magnitude of the force is F e(4.75 103 m/s) Bx . Since F 8.50 1016 N , Bx 8.50 1016 N 1.12 T . Bx 1.12 T . The sign of Bx is not determined by measuring (1.60 1019 C)(4.75 103 m/s) the magnitude of the force on the electron. B Bx2 Bz2 (1.12 T) (0.938 T) 2 1.46 T . tan Bz 0.938 T . 40° . B is in the xz-plane and is either at 40° from the +x-direction toward the Bx 1.12 T z-direction or 40° from the x-direction toward the z-direction . (b) B = Bx iˆ Bz kˆ . v = (3.2 km/s)( ˆj ) . F qv B (e)(3.2 km/s)( ˆj ) (Bx iˆ Bz kˆ ) e(3.2 103 m/s)(Bx (kˆ ) Bz iˆ ) . F = e(3.2 103 m/s)([1.12 T]kˆ [0.938 T]iˆ ) (4.80 1016 N)iˆ (5.73 1016 N)kˆ Fz 5.73 1016 N . 50.0° . The force is in the xz-plane and is Fx 4.80 1016 N directed at 50.0° from the x-axis toward either the +z or z axis, depending on the sign of Bx . F Fx2 Fz2 7.47 1016 N . tan 27.53. EVALUATE: If the direction of the force on the first electron were measured, then the sign of Bx would be determined. (a) IDENTIFY: Use Eq.(27.2) to relate v , B, and F . SET UP: The directions of v1 and F1 are shown in Figure 27.53a. F qv B says that F is perpendicular to v and B. The information given here means that B can have no z-component. Figure 27.53a The directions of v2 and F2 are shown in Figure 27.53b. F is perpendicular to v and B, so B can have no x-component. Figure 27.53b Both pieces of information taken together say that B is in the y-direction; B By ˆj. EXECUTE: Use the information given about F2 to calculate Fy : F2 F2iˆ, v2 v2kˆ, B By ˆj. F qv B says F iˆ qv B kˆ ˆj qv B (iˆ) and F qv B 2 2 2 2 y 2 y 2 2 y By F2 /(qv2 ) F2/(qv1 ). B has the maginitude F2 /(qv1 ) and is in the y-direction. (b) F1 qvB sin qv1 By / 2 F2 / 2 EVALUATE: v1 v2 . v2 is perpendicular to B whereas only the component of v1 perpendicular to B contributes to the force, so it is expected that F2 F1 , as we found. 27.55. IDENTIFY: The sum of the magnetic, electrical, and gravitational forces must be zero to aim at and hit the target. SET UP: The magnetic field must point to the left when viewed in the direction of the target for no net force. The net force is zero, so F FB FE mg 0 and qvB – qE – mg = 0. EXECUTE: Solving for B gives B qE mg (2500 106 C)(27.5 N/C) + (0.0050 kg)(9.80 m/s2 ) 3.7 T qv (2500106 C)(12.8 m/s) The direction should be perpendicular to the initial velocity of the coin. EVALUATE: This is a very strong magnetic field, but achievable in some labs.