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 find the radius of curvature of the path followed by the He+ ion in the magnetic field, we can use the equation for the centripetal force acting on a charged particle moving in a magnetic field.
The centripetal force is given by the equation:
F = qvB
Where:
F represents the force acting on the particle,
q represents the charge of the particle,
v represents the velocity of the particle, and
B represents the magnetic field strength.
Since the He+ ion has a charge of +2e (where e is the elementary charge), the force acting on it can be written as:
F = (2e)vB
The centripetal force required to make the ion follow a curved path can be equated to the force given by:
F = mv²/r
Where:
m represents the mass of the particle, and
r represents the radius of curvature.
Setting the two equations for force equal to each other, we have:
(mv²)/r = (2e)vB
We can further simplify the equation by substituting the mass of the helium nucleus as:
m = 4 × mass of a proton (since the helium nucleus contains two protons and two neutrons)
Substituting the values and rearranging the equation to solve for r, we get:
r = (4mv) / (2eB)
Now, let's fill in the given values:
m (mass of a proton) = 1.67 × 10^-27 kg
v (velocity) = speed of light (since it starts from rest and is given in vacuum)
e (elementary charge) = 1.6 × 10^-19 C
B (magnetic field strength) = 1.0 T
Plugging in these values, we can calculate the radius of curvature (r).