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A proton enters a region with a uniform electric field
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Chapter 30 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- 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_forwardAlong, straight wire lies along the z-axis and carries current I=60.0 A in the+z-direction. A small particle with mass 3.50 x 10 kg and charge 8.00 x 10 c is traveling in the vicinity of the wire. At an instant when the particle is on the y-axis at y 9.00 cm, its acceleration has components a-5.00 x 10³ m/s² and a 49.00 x 10³ m/s² Part A At that instant what are the 2- and y-components of the velocity of the particle? Express your answer in kilometers per second. Enter your answers numerically separated by a comma 19Η ΑΣΦ OPO Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Provide Feedback ? km/sarrow_forwardA proton moves through a uniform electric field given by vector E = 50.0 j hat V/m and a uniform magnetic field vector B = (0.200 i hat+ 0.300 j hat+ 0.400 k hat) T. Determine the acceleration of the proton when it has a velocity vector v = 200 i m/sarrow_forward
- A 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_forwardLet it be a particle with mass m = 2.35 × 10−3 kg and charge Q = 7.22 × 10−8 C. It has, at a certain moment, speed v = (4.00 × 105 ? / ?) j. Then it is influenced by a uniform magnetic field B = (1.52 ?) i + (0.68 ?) j. What is the a) modulus, b) direction and c) the direction of the acceleration suffered by the particle when in the influence of this field?arrow_forwardA particle with mass 0.8 gram and charge .02 μC is moving in the x-y plane with a velocity v(t) given by the formula v(t) = (15000 i + 30000 j )m/s. It enters a region of space where a magnetic field B exists, given by B = (1.5 i + 2j) T. What is the force on the charge? (Hint: i, j and k are the unit vectors along x, y and z-axis respectively. Use the relationships i x i = j x j = k x k = 0, and i x j = k, etc.) a) .0003 N along positive z b) .0003 N along positive x c) .0003 N along negative z. What is its acceleration? 0.375 m/s2 along negative z b) 0.375 m/s2 along positive x c) 0.375 m/s2 along positive zarrow_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 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 particle with positive charge q = 3.52 x 1018 C moves with a velocity v = (5î + 4ĵ – k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle, taking B = (3î + 2ĵ + k) T and E = (3î - j - 2k) V/m. (Give your answers in N for each component.) Ey = Ey = F, = (b) What angle does the force vector make with the positive x-axis? (Give your answer in degrees counterclockwise from the +x-axis.) ° counterclockwise from the +x-axis (c) What If? For what vector electric field would the total force on the particle be zero? (Give your answers in V/m for each component.) E, = V/m E, = V/m E, = V/marrow_forwardThis time I1 = 7.81, I2 = 3.46 A, and the two wires are separated by 9.30 cm. Now consider the charge q = 6.39 x 10^-6 C, located a distance of 9.33 cm to the right of wire I2, moving to the right at speed v = 63.7 m/s. What is the magnitude of the total magnetic force on this charge? A 1.13E-09 N B 6.43E-09 N C 3.22E-09 N D 1.61E-09 Narrow_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