A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.32 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?
3. A positive helium ion He+ is released from rest at the surface of a +1000 V flat electrode. It travels in the +x-direction (in vacuum) across a 1 mm gap, passes through a small hole in a +0 V electrode that is parallel to the first one, and enters a region of magnetic field. In the region with the magnetic field, it follows a path that curves in the +y-direction. Assume the magnetic field is uniform outside the electrodes, and zero in between them. For a field strength of B = 1.0 T, what is the radius of curvature of the resulting path in meters? A helium nucleus has two protons and two neutrons
To determine the magnitude of the field used for the electron, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field.
The formula for the magnetic force on a charged particle is given by:
F = qvB
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
In this case, we can use the given information that the proton and the electron move in the magnetic field with the same velocity and follow a circular path of the same radius. This means that the centripetal force acting on the proton and the electron must be the same, as it is responsible for keeping them in a circular path.
The centripetal force for a charged particle moving in a magnetic field is given by:
F = (mv²) / r
where m is the mass of the particle, v is the velocity, and r is the radius of the circular path.
Since the proton and the electron have the same velocity and follow a circular path with the same radius, we can equate the centripetal forces for the proton and the electron:
(q_proton * v * B_proton) = (q_electron * v * B_electron)
Since q_proton is the charge of a proton (e = 1.6 × 10^-19 C) and q_electron is the charge of an electron (-e = -1.6 × 10^-19 C), the charges have the same magnitude but opposite signs. We can eliminate v from both sides of the equation as well since it cancels out:
(q_proton * B_proton) = (q_electron * B_electron)
Plugging in the given value for the magnetic field used for the proton (B_proton = 0.32 T), we can now solve for B_electron:
(1.6 × 10^-19 C * 0.32 T) = (-1.6 × 10^-19 C * B_electron)
B_electron = - (1.6 × 10^-19 C * 0.32 T) / (-1.6 × 10^-19 C)
B_electron = 0.32 T
Therefore, the magnitude of the magnetic field used for the electron is 0.32 T.