A 3.25 g bullet picks up an electric charge of 1.85 μC as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 m/s, traveling perpendicular to the earth's magnetic field, which has a magnitude of 5.50 x 10-5 T. Part A Calculate the magnitude of the magnetic force on the bullet. Express your answer with the appropriate units. F = Submit Part B I a = μA Value Request Answer Calculate the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel. Express your answer with the appropriate units. μA Units Value ? Units ?

A 3.25 g bullet picks up an electric charge of 1.85 μC as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 m/s, traveling perpendicular to the earth's magnetic field, which has a magnitude of 5.50 x 10-5 T. Part A Calculate the magnitude of the magnetic force on the bullet. Express your answer with the appropriate units. F = Submit Part B I a = μA Value Request Answer Calculate the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel. Express your answer with the appropriate units. μA Units Value ? Units ?

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter28: Magnetic Fields

Section: Chapter Questions

Problem 46AP: Why is the following situation impossible? Figure P28.46 shows an experimental technique for...

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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?
(a) A cosmic ray proton moving toward the Earth at 5.00107m/s experiences a magnetic force of 1.701016N. What is the strength of the magnetic field it there is a 45° angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss.
One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 2.00 C is moving with a velocity of 150iMm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.
(a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat part (a), assuming the moving particle is an electron. Figure P19.2 Problems 2 and 22.
A proton travels with a speed of 3.00 106 m/s at an angle of 37.0 with the direction of a magnetic field of 0.300 T in the +y direction. What are (a) the magnitude of the magnetic force on the proton and (b) its acceleration?
An alpha-particle ( m=6.641027kg , q=3.21019C ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?
Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46