An á-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.16 106 V and then enters a uniform magnetic field whose magnitude is 3.00 T. The á-particle moves perpendicular to the magnetic field at all times.
(a) What is the speed of the á-particle?
1 m/s
(b) What is the magnitude of the magnetic force on it?
2 N
(c) What is the radius of its circular path?
3 m
To find the answers to these questions, we can use the principles of electrostatics, electromagnetism, and circular motion.
(a) To determine the speed of the alpha particle, we can use the principle of conservation of energy. The initial potential energy is converted into kinetic energy. The potential energy is given by the equation U = qV, where U is the potential energy, q is the charge, and V is the potential difference. The kinetic energy is given by the equation K = (1/2)mv^2, where m is the mass of the particle and v is its speed.
Setting the potential energy U equal to the kinetic energy K, we have:
qV = (1/2)mv^2
Rearranging the equation and plugging in the given values:
v = sqrt((2qV)/m) = sqrt((2 * 2e * 1.16e6 V) / 6.64e-27 kg)
≈ 507010 m/s
≈ 5.07 x 10^5 m/s
Therefore, the speed of the á-particle is approximately 5.07 x 10^5 m/s.
(b) The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvB, where F is the magnetic force, q is the charge, v is the velocity of the particle, and B is the magnetic field strength.
Plugging in the given values:
F = (2e)(5.07 x 10^5 m/s)(3.00 T)
≈ 3.04 x 10^(-13) N
Therefore, the magnitude of the magnetic force on the á-particle is approximately 3.04 x 10^(-13) N.
(c) For a charged particle moving in a uniform magnetic field, the path is circular, and the radius of the circular path can be determined using the equation r = mv/(qB), where r is the radius, m is the mass, v is the velocity, q is the charge, and B is the magnetic field strength.
Plugging in the given values:
r = (6.64 x 10^(-27) kg)(5.07 x 10^5 m/s) / (2e)(3.00 T)
≈ 2.21 x 10^(-4) m
≈ 0.221 mm
Therefore, the radius of the á-particle's circular path is approximately 0.221 mm.