CHAPTER 27 HOMEWORK SOLUTIONS
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1 CHAPTER 7 HOMEWORK SOLUTIONS 7.1. IDENTIFY and SET UP: Apply Eq.(7.) to calculate F. Use the cross products of unit vectors from Section EXECUTE: v m/siˆ m/s ˆj (a) B 1.40 Tˆ i F q v B C1.40 T m/s ˆˆ m/s ˆˆ i i j i iˆiˆ0, ˆj iˆk ˆ 8 4 ˆ F C 1.40 T m/s k Nkˆ EVALUATE: The directions of v and B are shown in 7.1a. The right-hand rule gives that v B is directed out of the paper (+z-direction). The charge is negative so F is opposite to v B ; 7.1a F is in the z- direction. This agrees with the direction calculated with unit vectors. (b) EXECUTE: B 1.40 Tk ˆ F q v B C1.40 T m/s ˆ ˆ m/s ˆ ˆ i k j k iˆkˆ ˆj, ˆjkˆ i ˆ F 4 ˆ 4 ˆ N N N ˆ j i N ˆ i j EVALUATE: The directions of v and B are shown in 7.1b. 7.1b The direction of F is opposite to v B since q is negative. The direction of F computed from the right-hand rule agrees qualitatively with the direction calculated with unit vectors.
2 7.8.IDENTIFY and SET UP: F= qvb= qb [ v ( iˆkˆ) v ( ˆj kˆ) v ( kˆk ˆ)] = qb [ v ( ˆj ) v ( iˆ)]. z x y z z x y EXECUTE: (a) Set the expression for F equal to the given value of F to obtain: 7 Fy ( N) vx 106 m s 9 qbz ( C)( 1.5 T) 7 Fx ( N) vy 48.6 m s. 9 qb ( C)( 1.5 T) z (b) vz does not contribute to the force, so is not determined by a measurement of F. Fy Fx (c) vf vf x x vf y y vf z z Fx Fy 0; 90. qbz qbz EVALUATE: The force is perpendicular to both v and B, so B F is also zero IDENTIFY: When B is uniform across the surface, B B A BAcos. SET UP: A is normal to the surface and is directed outward from the enclosed volume. For surface abcd, A= Aiˆ. For surface befc, A= Akˆ. For surface aefd, cos 3/5 and the flux is positive. EXECUTE: (a) ( abcd B ) B A 0. (b) ( befc B ) B A (0.18 T)(0.300 m)(0.300 m) Wb. 3 (c) ( aefd) B A BAcos (0.18 T)(0.500 m)(0.300 m) Wb. B 5 (d) The net flux through the rest of the surfaces is zero since they are parallel to the x-axis. The total flux is the sum of all parts above, which is zero. EVALUATE: The total flux through any closed surface, that encloses a volume, is zero. 7.. IDENTIFY: For motion in an arc of a circle, a and the net force is radially inward, toward the R center of the circle. 7 SET UP: The direction of the force is shown in 7.. The mass of a proton is kg. v EXECUTE: (a) F is opposite to the right-hand rule direction, so the charge is negative. F ma gives 19 v qbr 3( C)(0.50 T)(0.475 m) 6 qvbsin m. 90 and v.8410 m/s. R 7 m 1( kg) (b) w F q vbsin 3( C)(.8410 m/s)(0.50 T)sin N. B 7 5 1( kg)(9.80 m/s ) 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. 7.
3 7.39. IDENTIFY and SET UP: The magnetic force is given by Eq.(7.19). FI when the bar is just ready to levitate. When I becomes larger, FI and FI is the net force that accelerates the bar upward. Use Newton's nd law to find the acceleration kg9.80 m/s (a) EXECUTE: IlB, I 3.67 A lb m T E IR 3.67 A V (b) R.0, I E / R816.7 V / A FI IlB 9 N a F / m 113 m/s I EVALUATE: I increases by over an order of magnitude when R changes to FI and a is an order of magnitude larger than g IDENTIFY: IABsin, where is the angle between B and the normal to the loop. SET UP: The coil as viewed along the axis of rotation is shown in 7.44a for its original position and in 7.44b after it has rotated EXECUTE: (a) The forces on each side of the coil are shown in 7.44a. F1F 0 and F3 F4 0. The net force on the coil is zero. 0 and sin 0, so 0. The forces on the coil produce no torque. (b) The net force is still zero and the net torque is (1)(1.40 A)(0.0 m)(0.350 m)(1.50 T)sin N m. The net torque is clockwise in 7.44b and is directed so as to increase the angle. EVALUATE: For any current loop in a uniform magnetic field the net force on the loop is zero. The torque on the loop depends on the orientation of the plane of the loop relative to the magnetic field direction. 7.44
4 7.57. (a) IDENTIFY and SET UP: The maximum radius of the orbit determines the maximum speed v of the protons. Use Newton's nd law and a c v / R for circular motion to relate the variables. The energy of 1 the particle is the kinetic energy K mv. EXECUTE: F = ma gives qvb mv ( / R) 19 qbr ( C)(0.85 T)(0.40 m) 7 v m/s. The kinetic energy of a proton moving 7 m kg with this speed is K mv ( kg)( m/s) J 5.6 MeV R (0.40 m) 8 (b) The time for one revolution is the period T s 7 v m/s (c) qbr q BR Km K mv m. Or, B. m m q R B is proportional to K, so if K is increased by a factor of then B must be increased by a factor of. B (0.85 T) 1. T. 19 qbr (3.010 C)(0.85 T)(0.40 m) 7 (d) v m/s 7 m kg K mv ( kg)( m/s) J 5.5 MeV, the same as the maximum energy for protons. EVALUATE: We can see that the maximum energy must be approximately the same as follows: From qbr 1 part (c), K m. For alpha particles q is larger by a factor of and m is larger by a factor m of 4 (approximately). Thus q / m is unchanged and K is the same.
5 7.67. IDENTIFY: The force exerted by the magnetic field is given by Eq.(7.19). The net force on the wire must be zero. SET UP: For the wire to remain at rest the force exerted on it by the magnetic field must have a component directed up the incline. To produce a force in this direction, the current in the wire must be directed from right to left in 7.61 in the textbook. Or, viewing the wire from its left-hand end the directions are shown in 7.67a. 7.67a The free-body diagram for the wire is given in 7.67b. EXECUTE: F 0 FI cos Mgsin 0 FI ILBsin 90 since B is perpendicular to the current direction. y 7.67b Mg tan Thus (ILB) cos Mg sin 0 and I LB EVALUATE: The magnetic and gravitational forces are in perpendicular directions so their components parallel to the incline involve different trig functions. As the tilt angle increases there is a larger component of Mg down the incline and the component of F up the incline is smaller; I must increase with to compensate. As 0, I 0 and as 90, I. I
6 7.75. IDENTIFY: For the loop to be in equilibrium the net torque on it must be zero. Use Eq.(7.6) to calculate the torque due to the magnetic field and use Eq.(10.3) for the torque due to the gravity force. SET UP: See 7.75a. Use A 0, where point A is at the origin. 7.75a EXECUTE: See 7.75b. r sin (0.400 m)sin 30.0 The torque is clockwise; directed into the paper. Figu re 7.75 b For the loop to be in equilibrium the torque due to B must be counterclockwise (opposite to ) and it must be that. See 7.75c. B 7.75c B Bsin IABsin 60.0 B gives IABsin m sin 30.0 m B B. For this torque to be counterclockwise ( B directed out of the paper), B must be in the y-direction g/cm 8.00 cm 6.00 cm 4. g kg msin 30.0 A m m m B B 3 IAsin kg 9.80 m/s m sin T 3 (8. A)( m )sin 60.0 As the loop swings up the torque due to B decreases to zero and the torque due to EVALUATE: increases from zero, so there must be an orientation of the loop where the net torque is zero. is
7 mv IDENTIFY and SET UP: In the magnetic field, R. Once the particle exits the field it travels in a qb straight line. Throughout the motion the speed of the particle is constant mv ( kg)( m/s) EXECUTE: (a) R 5.14 m. qb 6 ( C)(0.40 T) (b) See The distance along the curve, d, is given by d R m sin, so 5.14 m rad. d Rθ (5.14 m)( rad) 0.5 m. And d 0.5 m 6 t s. 5 v m/s 3 (c) x1 dtan( θ / ) (0.5 m)tan (.79 / ) m. (d) x x1 x, where x is the horizontal displacement of the particle from where it exits the field region to where it hits the wall. x (0.50 m)tan m. Therefore, 3 x m m m. EVALUATE: d is much less than R, so the horizontal deflection of the particle is much smaller than the distance it travels in the y-direction. 7.89
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