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 helium ion (He+) in the magnetic field, we can use the formula for the radius of curvature of a charged particle moving in a magnetic field.
The formula for the radius of curvature (r) is given by:
r = (m*v) / (q*B)
Where:
- m is the mass of the particle
- v is the velocity of the particle
- q is the charge of the particle
- B is the magnetic field strength
Let's break down the problem step by step to find the values we need to plug into the formula:
1. Charge of the helium ion (He+):
The helium ion (He+) has a charge of +2e, where e is the elementary charge.
Therefore, q = +2 * e
2. Mass of the helium ion (He+):
The helium ion consists of two protons and two neutrons, so its mass (m) can be calculated as follows:
m = (2 * mass of a proton) + (2 * mass of a neutron)
= (2 * mass of a proton) + (2 * (mass of a proton + mass of a neutron))
The mass of a proton is approximately 1.673 x 10^(-27) kg.
The mass of a neutron is approximately 1.675 x 10^(-27) kg.
Plugging in these values, we get:
m = (2 * 1.673 x 10^(-27) kg) + (2 * (1.673 x 10^(-27) kg + 1.675 x 10^(-27) kg))
3. Velocity of the helium ion (He+):
The problem does not mention the initial velocity of the helium ion, but it does state that it is released from rest. Therefore, the initial velocity is zero (v = 0 m/s).
4. Magnetic field strength (B):
The problem states that the magnetic field strength (B) is 1.0 T (Tesla).
Now, with all the values determined, let's substitute them into the formula to find the radius of curvature (r):
r = (m * v) / (q * B)
Remember that v = 0 m/s because the helium ion is released from rest.
After plugging in all the values, you can calculate the radius of curvature (r) in meters.