Phys 201Review problems-Ch29 Magnetic FieldsDr.Hikmat A. Hamad

1. An electron has a velocity of 6.0 106 m/s in the positive x direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T. What is the magnitude of the acceleration of the electron at this point? Ans: 2.6 1018 m/s2

2. A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 1027 kg, q = 1.6 1019 C). Ans: 13 mm

3. A particle (q = 4.0 C, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a direction that is 50 away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s2. What is the magnitude of the magnetic field? Ans: 4.7 mT

4. A 2.0-C charge moves with a velocity of (2.0i + 4.0j + 6.0k) m/s and experiences a magnetic force of (4.0i 20j + 12k) N. The x component of the magnetic field is equal to zero. Determine the z component of the magnetic field.. Ans: +5.0 T

5. A particle (mass = 2.0 mg, charge = 6.0 C) moves in the positive direction along the x axis with a velocityof 3.0 km/s. It enters a magnetic field of (2.0i + 3.0j + 4.0k) mT. What is the acceleration of the particle? Ans: (36j 27k) m/s2

6. A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits? Ans: 1.3

7. A segment of wire carries a current of 25 A along the x axis from x = 2.0 m to x = 0 and then along the y axis from y = 0 to y = 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z- direction. What is the magnitude of the force on this segment of wire? Ans: 3.6 N

8. A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60km/s enters this field. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine theangle ( ) between the magnetic field and the velocity of the proton. Ans: 51 9. A particle (m = 3.0 g, q = 5.0 C) moves in a uniform magnetic field given by (60j) mT. At t = 0 the velocity of the particle is equal to (30j 40k) m/s. What is the subsequent path of the particle ? Ans: helical with a 40-cm radius.

10. A charged particle moves in a region of uniform magnetic field along a helical path (radius = 5.0 cm, pitch = 12 cm, period = 5.0 ms). What is the speed of this particle as it moves along this path? Ans: 67 m/s

11. What is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T? Ans: 9.9 mm

12. An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 T. Theelectron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm. What is the speed of thiselectron as it moves in this region? Ans: 35 km/s

13. A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularlythrough a uniform 0.60-T magnetic field. What is the radius of the resulting path? Ans: 12 mm

14. A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to selectparticles of various energies. If a magnetic field of magnitude B is used to select a particle of a certain energyand mass, what magnitude of magnetic field is needed to select a particle of equal mass but twice the energy?Ans:0.71 B15. A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails (Fig. P29.37) that are d = 12.0 cmapart and L = 45.0 cm long. The rod carries a current of I = 48.0 A in the direction shown and rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails? Ans:1.07 m/s(Hint: use W=E = 1/2mv2 +1/2 I2 where I =1/2mR2 and =v/R)

16. An electron with a kinetic energy of 22.5 eV moves into a region of uniform magnetic field of magnitude 4.55 x104 T. The angle between the directions of and the electrons velocity is 65.5. What is the pitch (p) of the helical path taken by the electron? Ans: 9.16 cmPhy201 Review Problems Ch29Page 1