1 Chapter 22: Magnetism 11. A charged particle moves in a region in which a magnetic field exists. Solve the magnetic force equation (equation 22-1) for the angle θ that would produce the specified force. 1. (a) Solve equation 22-1 for θ : 1sinFqvBθ−= 2. Insert the numerical values for F = 4.8 µN: ( )( )( )6164.8 10 Nsin 810.32 10 C 16 m/s 0.95 Tθ−−−×= = °× 3. (b) Insert the numerical values for F = 3.0 µN: ( )( )( )6163.0 10 Nsin0.32 10 C 16 m/s 0.95 Tθ−−−×= = 38°× 4. (c) Insert the numerical values for F = 0.10 µN: ( )( )( )7161.0 10 Nsin 1.20.32 10 C 16 m/s 0.95 Tθ−−−×= = °× 12. A charged particle moves in a region in which a magnetic field exists. Use a ratio together with equation 22-1 to determine the force the particle experiences after changing its speed and the angle its velocity makes with the magnetic field. 1. Use equation 22-1 to make a ratio: ( )( )new new new new newold old old old old6.3 m/s sin 25sin sin0.099sin sin 27 m/s sin90F qv B vF qv B vθθθθ°= = = =° 2. Now solve for newF: ( )( )45new old0.099 0.099 2.2 10 N 2.2 10 NFF−−= = ×=× 13. An ion moves with constant speed in a magnetic field. The ion experiences no magnetic force when it is moving in the ˆydirection, so we conclude that the magnetic field also points in the ˆydirection. When the ion travels in the xy plane and along the line y = x, it moves at an angle of 45° with respect to .B When it moves in the ˆx direction, it experiences the maximum magnetic force. Apply equation 22-2, letting maxF q vB=: ( )16 16maxsin 45 6.2 10 N sin 45 4.4 10 NFF−−= °= × °= × 16. The directions of the electric field, magnetic field, velocity, and force vectors involved in this problem are depicted in the diagram at the right. The particle is positively charged, so the electric field E exerts a force in the ˆ+x direction. Because the net force acting on the particle is in the ˆ+x direction, the magnetic force must also be along the ˆx direction, but it could be in either the ˆ+x direction or the ˆ−x direction. We can compare the relative magnitudes of the net force and the electric force to decide in which of these two directions the magnetic force must point. Once the direction of BF is known, the Right-Hand Rule can be used to discern the direction of the particle’s velocity .v 1. Find EF using equation 19-9: ( )( )( )63Eˆ ˆˆ6.60 10 C 1250 N/C 8.25 10 NqE−−==×=×Fx x x 2. Because net EFF<, the force due to the magnetic field opposes the force due to the electric field and must point in the ˆ−x direction. According to the RHR, v is in the negative y-direction.2 3. Write Newton's Second Law in the x direction and solve the expression for v: net3net61 1 N 6.23 10 N1250 1.02 T C6.60 10 C300 m/s 0.30 km/sxF qE qvBqvB qE FFvEBq−−= −= −×=−= −×= =∑ 22. A velocity selector is constructed by forming perpendicular and EB fields as indicated in the diagram at the right. According to the Left-Hand Rule (or to the opposite of the Right-Hand Rule), the magnetic force on the negatively charged particle is in the ˆ−y direction. The force due to the electric field must therefore point in the ˆ+y direction in order to oppose the magnetic force, and E must point in the ˆ−y direction because the charge is negative. The magnitude of E must satisfy the relation ,v EB= as discussed in the text. Solve v EB= for E and incorporate the direction of E as discussed in the Strategy: ( )( )( )( )( )33ˆˆ4.5 10 m/s 0.96 Tˆ4.3 10 N/CvB= −= × −=−×Ey yy 25. The magnetic force on a charged particle traveling at constant speed causes it to move in a circle. The magnetic force provides the centripetal force required to keep the particle moving in a circle. The radius of the circle in terms of m, v, q, and B is given by equation 22-3. We must first solve equation 22-3 for the speed v of the particle, then find the time it takes the particle to complete one orbit by dividing the circumference by the speed. 1. (a) Solve equation 22-3 for v: ( )( )( )6512.5 10 C 21.8 m 1.01 T2.80 10 kg9.83 m/sqrBvm−−×= =×= 2. (b) Divide the circum- ference by the speed: ( )( )2 21.8 m2213.9 s9.83 m/srmtv qBπππ= = = = 26. The magnetic force on a charged particle traveling at constant speed causes it to move in a circle. Apply the Right-Hand Rule to the diagram at the right in order to determine whether the particle is positively or negatively charged. Then use equation 22-3, which gives the radius of the circle in terms of m, v, q, and B, in order to find the mass m of the particle. 1. (a) According to the RHR, a positively charged particle would experience a force to the left. Because the particle is experiencing a force to the right, it must be negatively charged. 2. (b) Solve equation 22-3 for m: ( )( )( )( )192761.60 10 C 0.520 m 0.180 T1.0 u1.5 u1.67 10 kg6.0 10 m/serBmv−−×== ×=××3 34. A current-carrying wire experiences a force due to the presence of a magnetic field. Solve equation 22-4 for B to answer the question of part (a), and solve the same equation for θ to answer the question of part (b). 1. (a) Write equation 22-4 in terms of force per unit length: sinFIBLθ= 2. Solve for B: ( )0.033 N/m0.041 T 41 mTsin 6.2 A sin7.5FLBIθ= = = =° 2. (b) Solve equation 22-4 for θ : ( )( )110.015 N/msin sin 3.46.2 A 0.041 TFLIBθ−−= = = ° 35. A wire carries a current in a region where the magnetic field exerts an upward force on the wire. Set the magnitude of the magnetic force (equation 22-4) equal to the weight of the wire to find the current required to levitate the wire. The force is maximum when the current is perpendicular to the field. Therefore, the minimum required current occurs in the perpendicular configuration. 1. Set the magnetic force equal to the weight: sin90F mg ILB ILB= = °= 2. Solve for I: ( )( )( )( )20.75 kg 9.81 m/s2.4 A3.6 m 0.84 TmgILB= = = 43. Two wire loops have the same perimeter length, but one has the shape of a circle and the other a square. Each loop carries the same current and is immersed in the same magnetic field. The two loops each experience a torque due to the magnetic field. The maximum torque is proportional to the area of the loop (equation 22-5).