Solve the following:(a) An alpha particle (m=6.64 x 10-27 kg) contains two protons and two neutrons has no electrons to balancethe positive charge of the two protons, hence it has a charge of +2. (Given charge of a proton is 1.6 x 10-19C).Assume one such alpha particle travels in a circular path of radius 33 cm in a uniform magnetic field ofmagnitude 1.75T.) What is the speed of the alpha particle ?(b) If a proton of (m=1.67x10-27kg) travels with the same speed as of the alpha particle in the same magneticfield,() Calculate the radius of the circular path traced by the proton?(ii) Also, calculate the ratio of the radius of the alpha particle to the radius of the proton?

Answered: Image /qna-images/answer/738076e4-2482-4f83-bd98-9c6d40ce79d3.jpg

Solve the following: (a) An alpha particle (m=6.64 x 10-27 kg) contains two protons and two neutrons has no electrons to balance the positive charge of the two protons, hence it has a charge of +2. (Given charge of a proton is 1.6 x 10-19C). Assume one such alpha particle travels in a circular path of radius 33 cm in a uniform magnetic field of magnitude 1.75T. (i) What is the speed of the alpha particle ? (b) If a proton of (m=1.67x10-27kg) travels with the same speed as of the alpha particle in the same magnetic field, (i) Calculate the radius of the circular path traced by the proton? (ii) Also, calculate the ratio of the radius of the alpha particle to the radius of the proton?

Solve the following: (a) An alpha particle (m=6.64 x 10-27 kg) contains two protons and two neutrons has no electrons to balance the positive charge of the two protons, hence it has a charge of +2. (Given charge of a proton is 1.6 x 10-19C). Assume one such alpha particle travels in a circular path of radius 33 cm in a uniform magnetic field of magnitude 1.75T. (i) What is the speed of the alpha particle ? (b) If a proton of (m=1.67x10-27kg) travels with the same speed as of the alpha particle in the same magnetic field, (i) Calculate the radius of the circular path traced by the proton? (ii) Also, calculate the ratio of the radius of the alpha particle to the radius of the proton?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

An electron is accelerated through 2.70 103 V from rest and then enters a uniform 2.80-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience? (b) What is the minimum magnitude of the magnetic force this particle can experience?
RP
An electron is accelerated from rest by a potential difference of 388 V. It then enters a uniform magnetic field of magnitude 180 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field. (a) Number (b) Number UL Units Units
In the Bohr model of the atom, electrons make circular orbits around the nucleus. Consider an atom where an electron is in orbit with a radius of 5.90 ✕ 10−11 m and a speed of 2.20 ✕ 106 m/s. (a) What is the magnitude of the magnetic dipole moment (in A · m2) due to the electron's motion? A · m2 (b) If the electron moves in a horizontal circle, counterclockwise as seen from above, what is the direction of this magnetic dipole moment vector? downwardupward to the leftto the right
(a) An alpha particle (m=6.64 x 10-27 kg) contains two protons and two neutrons has no electrons to balance the positive charge of the two protons, hence it has a charge of +2. (Given charge of a proton is 1.6 x 10-19C). Assume one such alpha particle travels in a circular path of radius 28 cm in a uniform magnetic field of magnitude 2.25T. (i) What is the speed of the alpha particle ? (b) If an electron of (m=9.11x10-3'kg) travels with the same speed as of the alpha particle in the same magnetic field, (1) Calculate the radius of the circular path traced by the electron? (ii) Also, calculate the ratio of the radius of the electron to the radius of the alpha particle?
1) A proton passes through a uniform magnetic field of magnitude 4.0 μT at a speed of 6.43x10^4 m/s. (a) What is the magnitude of maximum magnetic force it can experience? (b) If the proton is replaced by the electron, will it experience the same maximum force? Complete solution.
(a) An electron moves through a uniform magnetic field B=B_x i +(5B_x)j with velocity V= (2i+2j)m/s experienced a force (5x10^-19 N)k. What would be the value of B_x?
What uniform magnetic field, applied perpendicular to a beam of electrons moving at 2.70 ✕ 106 m/s, is required to make the electrons travel in a circular arc of radius 0.400 m?
An electron initially at rest is accelerated through a potential difference of 3 kV in the positive x- direction. It enters a region where a uniform magnetic field of 5.0-mT is in a positive y-direction (perpendicular to the velocity of the electron). Calculate the radius of the path this electron will follow in the magnetic field. (e = 1.60 × 10-19 c, m electron = 9.11 × 10¯31 kg)
An electron that has velocity т i = (2.0 x 106)i+ (3.0 × 106 10", %3D moves through the uniform magnetic field B = (0.030 T)î – (0.15T)f. (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.
(a) At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.85 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.20 ✕ 10−5 T? m/s (b) What would the radius (in m) of the path be if the proton had the same speed as the electron? m (c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? m
A wire is formed into a circle having a diameter of 10.2 cm and is placed in a uniform magnetic field of 2.77 mT. The wire carries a current of 5.00 A. Find the maximum torque on the wire.