# 17.pdf

## Physics 261 with Mary Ewell at George Mason University *

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CHAPTER 27 – Magnetism 1. (a) The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have F max = ILB, or F max /L = IB = (7.40 A)(0.90 T) = 6.7 N/m. (b) We find the force per unit length from F/L = IB sin 45.0° = (F max /L) sin 45.0° = (6.7 N/m) sin 45.0° = 4.7 N/m. 10. We find the force per unit length from F/L = I(i × B) = (3.0 A)[i × (0.20i – 0.30 j + 0.25k)T] = (3.0 A)(– 0.30k – 0.25j)T = (– 0.75j – 0.90k) N/m = (– 7.5 × 10 –3 j – 9.0 × 10 –3 k) N/cm. 11. We choose the coordinate system shown in the diagram. We select a differential element of the curved wire, y x d¬ dF I a b B d¬ = dx i + dy j, on which the force is dF = I d¬ × B = I(dx i + dy j) × (– Bk) = IB(dx j – dy i). We find the resultant force by integration: F = ∫ dF = IB ∫ (dx j – dy i) = IB(Δx j – Δy i), where Δx = x b – x a , and Δy = y b – y a . If we have the same current in the straight wire, the resultant force is F = I(Δx i + Δy j) × (– Bk) = IB(Δx j – Δy i), which is the same result. 14. To find the direction of the force on the negative charge, we point our fingers in the direction of v and curl them into the magnetic field B. Our thumb points in the direction of the force on a positive charge. Thus the force on the negative charge is opposite to our thumb. (a) Fingers out, curl down, thumb right, force left. (b) Fingers down, curl back, thumb right, force left. (c) Fingers in, curl right, thumb down, force up. (d) Fingers right, curl up, thumb out, force in. (e) Fingers left, but cannot curl into B, so force is zero. (f) Fingers left, curl out, thumb up, force down. 15. We assume that we want the direction of B that produces the maximum force, i. e., perpendicular to v. Because the charge is positive, we point our thumb in the direction of F and our fingers in the direction of v. To find the direction of B, we note which way we should curl our fingers, which will be the direction of the magnetic field B. (a) Thumb out, fingers left, curl down. (b) Thumb up, fingers right, curl in. (c) Thumb down, fingers in, curl right. 21. The magnetic force provides the radial acceleration, so we have F = evB = mv 2 /r, so r = mv/eB = (2mK) 1/2 /eB = [2(1.67 × 10 –27 kg)(5.0 × 10 6 eV)(1.60 × 10 –19 J/eV)] 1/2 /(1.60 × 10 –19 C)(0.20 T) = 1.6 m. 24. The magnetic force produces an acceleration perpendicular to the original motion: a ⊥ = qvB/m = (13.5 × 10 –9 C)(160 m/s)(5.00 × 10 –5 T)/(3.40 × 10 –3 kg) = 3.18 × 10 –8 m/s 2 . The time the bullet takes to travel 1.00 km is t = L/v = (1.00 × 10 3 m)/(160 m/s) = 6.25 s. The small acceleration will produce a small deflection, so we assume the perpendicular acceleration is constant in magnitude and direction. We find the deflection of the electron from y = v 0y t + !at 2 = 0 + !(3.18 × 10 –8 m/s 2 )(6.25 s) 2 = 6.20 × 10 –7 m. This justifies our assumption of constant acceleration. 25. If we assume the positively-charged ion is traveling west around the equator, the magnetic force will be down. This force and mg will produce the radial acceleration: mg + evB = mv 2 /r, or v 2 – (eBr/m)v – gr = 0. We evaluate the constants: eBr/m = (1.60 × 10 –19 C)(0.40 × 10 –4 T)(6.38 × 10 6 m + 5.0 × 10 3 m)/(238 u)(1.66 × 10 –27 kg/u) = 1.034 × 10 8 m/s; gr = (9.80 m/s 2 )(6.38 × 10 6 m + 5.0 × 10 3 m) = 6.25 × 10 7 m 2 /s 2 . Thus we have v 2 – (1.034 × 10 8 m/s)v – 6.25 × 10 7 m 2 /s 2 = 0. The solutions to this quadratic equation are + 1.034 × 10 8 m/s, – 0.605 m/s. We see that for the large positive velocity (to the west) the mg term is negligible, so the magnetic force provides the radial acceleration. For the small negative velocity (to the east), the radial acceleration is negligible because the radius is so large; the upward magnetic force essentially balances the mg force. Thus the appropriate answer is 1.034 × 10 8 m/s (west) and gravity can be ignored. Note that this speed is close to c, so a relativistic correction is necessary. 32. When the loop is parallel to the magnetic field, the torque is maximum, so we have τ = NIAB; 0.185 m · N = (1)(7.10 A)π(0.0650) 2 B, which gives B = 1.96 T. 34. The angular momentum of the electron for the circular orbit is L = mvr. The time for the electron to go once around the orbit is T = 2πr/v, so the effective current is I = e/T = ev/2πr. The magnetic dipole moment is M = IA = (ev/2πr)πr 2 = evr/2 = (e/2m)L. 35. (a) The angle between the normal to the coil and the field is 24.0°, so the torque is τ = NIAB sin θ = (12)(7.10 A)π(0.0850 m) 2 (5.50 × 10 –5 T) sin 24.0° = 4.33 × 10 –5 m · N. (b) From the directions of the forces shown on the diagram, the north edge of the coil will rise. 57. We assume the magnetic field makes an angle θ with the vertical. From a view along the rod, we have the forces shown in the diagram. For the z-direction we have F B sin θ + F N – mg = 0, or F N = mg – ILB sin θ. To start motion in the x-direction, we have F B cos θ > F fr,max = μ s F N . When we combine the two equations, we get B > μ s mg/IL(cos θ + μ s sin θ). To find the angle for the minimum field, we find the angle that will make the denominator maximum by differentiating: d(cos θ + μ s sin θ)/dθ = – sin θ + μ s cos θ = 0, or tan θ = μ s = 0.50, θ = 26.6°. Thus B > (0.50)(0.40 kg)(9.80 m/s 2 )/(40 A)(0.22 m)(cos 26.6° + 0.50 sin 26.6°) = 0.20 T. θ B Earth I (west) F north I (east) F up B I z θ mg F fr θ x F N F B The minimum magnetic field is 26.6° from the vertical. 61. If the beam is perpendicular to the magnetic field, the force from B R L θ v v d the magnetic field is always perpendicular to the velocity, so it will change the direction of the velocity, but not its magnitude. The radius of the path in the magnetic field is R = mv/qB. Protons with different speeds will have paths of different radii. Thus slower protons will deflect more, and faster protons will deflect less, than those with the design speed. We find the radius of the path from R = mv/qB = (1.67 × 10 –27 kg)(0.75 × 10 7 m/s)/(1.60 × 10 –19 C)(0.33 T) = 0.237 m. Because the exit velocity is perpendicular to the radial line from the center of curvature, the exit angle is also the angle the radial line makes with the boundary of the field: sin θ = L/R = (0.050 m)/(0.237 m) = 0.211, so θ = 12°. Neil Goldman CHAPTER 27
Magnetism

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