Concept explainers
A proton enters a region with a uniform electric field
Trending nowThis is a popular solution!
Chapter 30 Solutions
Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term
- A particle moving downward at a speed of 6.0106 m/s enters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If B = 0.25 T and the charge-to-mass ratio (q/m) of the particle is 40107 C/kg. what is ±e radius at the path? (c) What is the speed of the particle after c has moved in the field for 1.0105s ? for 2.0s?arrow_forwardt1, the proton has a velocity given by v = (vz î + (2000m/s)ĵ) and the magnetic force of 2.3: A proton moves through a uniform magnetic field given by B = (30 î + 20ĵ) mT. At a time = (4 * 10-17 N) k. At that instant, what is the velocity vx? the proton is FB =arrow_forwardA particle of mass m and charge q is accelerated along the +x axis (in the plane of the page) from rest through an electric potential difference V. The particle then enters a region, defined by x> 0, containing a uniform magnetic field. V = 400.0 V B = 0.850 T q = -1.602 × 10-19 m = 6.68 × 10-27 kg B = 0 9. What is the radius of the particle's path, and how long does it take the particle to exit the magnetic field region?arrow_forward
- A long, straight wire lies along the r-axis and carries current I = 60.0 A in the +x-direction. A small particle with mass 3.00 x 10-6 kg and charge 8.00 × 10-3 C is traveling in the vicinity of the wire. At an instant when the particle is on the y-axis at y = 7.00 cm, its acceleration has components a, = -5.00 x 10³ m/s? and a, = 19.00 x 103 m/s². Part A At that instant what are the - and y-components of the velocity of the particle? Express your answer in kilometers per second. Enter your answers numerically separated by a comma. Vz, Vy = km/s Submit Request Answerarrow_forwardA particle with charge 4.16 x 10^-6 C moves at 5.78 x 10^6 m/s through a magnetic field of strength 2.58 T. The angle between the particle s velocity and the magnetic field direction is 35.1 degrees and the particle undergoes an acceleration of 11.9 m/s^2. What is the particle s mass? 17.9 kg 3.00 kg 6.0 kg 7.1 kgarrow_forwardA particle with charge q = 1.2 C and a long wire carrying current I = 6 A is on the xy-plane at the moment the charge is at point P at distance d = 0.4 cm from the wire and has velocity v = 1.8 m/s in the I d x-direction. What is the magnitude of the magnetic force acting on the charge in units of nanonewtons? (u = 47 x 10-7T · m/A) Answer:arrow_forward
- A long, straight wire lies along the x-axis and carries current I = 60.0 A in the +x-direction. A small particle with mass 3.50 x 100 kg and charge 8.00 x 10-3 C is traveling in the vicinity of -6 the wire. At an instant when the particle is on the y-axis at y=7.00 cm, its acceleration has components a = -5.00 x 103 m/s² and ay=+9.00 x 103 m/s².arrow_forwardA particle with mass 3×10−2 kgkg and charge +7 μCμC enters a region of space where there is a magnetic field of 1 TT that is perpendicular to the velocity of the particle. When the particle encounters the magnetic field, it experiences an acceleration of 17 m/s2m/s2 . What is the speed of the particle when it enters the magnetic-field region? Express your answer in meters per second.arrow_forwardA positively charged particle slides off a frictionless and smooth surface with initial X X X X X X X X speed v, =v, . It then enters into a region with X X X X X x x x X uniform magnetic field pointing into the page. X X X X x x x x X x x x x x X x x x The particle eventually hits the ground with final speed v, air time and range S · The particle slides off the edge again with identical initial speed, but this time without the magnetic field. The particle then hits the ground with final speed v,, air time t, , and range s,. Determine the relationships between v, and v,, t, and t, , as well as s, and s, (3 marks). Please note that no calculation is needed. You were to provide the relationship in the sense that whether one variable is greater than, smaller than, or equals to the other one (>, <, or =). Provide a rough explanation (3 marks).arrow_forward
- A proton moves in a helical path at speed v = 5.10 x 107 m/s high above the atmosphere, where Earth's magnetic field has magnitude B = 3.80 x 10-6 T. The proton's velocity makes an angle of 25.0° with the magnetic field. The mass of the proton is 1.673x10-27 kg. Find the radius of the helix. kmarrow_forwardA charged particle enters a region with uniform magnetic field and leaves in a perpendicular with respect to initial direction. Find the sign and the speed of the charged particle is B=3,22 T, m=7,17x10 21 kg, q=8e and the distance travelled in the magnetic field is I=4,11 cm, where e is charge of one electron. IX X x X XI B ix x X x a. positive, 1504,12 cm/s b. negative, 1128,09 cm/s c. positive, 752,06 cm/s d. negative, 1504,12 cm/s e. positive, 1128,09 cm/sarrow_forwardThe plane of a conducting wire of the form carrying a current l=2A is perpendicular to a uniform magnetic field of magnitude B=0.6 T. The lengths of the straight parts of the current carrying conductor wire located in the magnetic field are l3=12 =40 cm and the radius of the semicircular arc is 18 cm. Express the net magnetic force acting on the current carrying conductor in unit vectors. a) F = 0,91 j- 0.48 i b) F = 1,52 j - 0,76 i F = 1,46 j-0,73 i F=0,46 j – 0,87 iarrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning