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
assuming the "flat" electrode produces a uniform E field, then E=1000/.001 volts/meter
So force=E*q*distance=mass*accleration.
Vf^2=Vi^2+2ad=0+2(Eqd/m)d you know gap d, E, q (He is +2e, mass is 4xprotonmass)
solve for vf. That is the velocity leaving the hole.
Now in the magnetic field
forcemagnetic=mv^2/r
Bqv=mv^2/r solve for r.
To find the radius of curvature of the path followed by the helium ion (He+) in the magnetic field, we can use the Lorentz force equation. The Lorentz force is given by the formula:
F = q(v x B)
where F is the force experienced by the charged particle, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field vector.
In this case, the helium ion is positively charged with a charge of +2e (where e is the elementary charge), and it starts from rest. Initially, the velocity (v) of the helium ion is zero.
When a charged particle enters a magnetic field, it experiences a force perpendicular to both the velocity vector and the magnetic field vector. This force causes the particle to follow a curved path.
The magnitude of the force can be written as:
F = |q|(v)(B)
Since the velocity of the helium ion is initially zero, we need to consider its final velocity just before it enters the magnetic field. The electrostatic potential difference between the two electrodes will cause an acceleration in the helium ion, eventually giving it a final velocity.
The electric potential difference (V) between the two electrodes can be given as:
V = ΔE/q
where ΔE is the change in energy of the particle and q is its charge. In this case, the electric potential difference is given as +1000 V.
The change in energy (ΔE) can be expressed as:
ΔE = KE_final - KE_initial
Since the helium ion starts from rest, its initial kinetic energy is zero. Therefore, the change in energy simplifies to:
ΔE = KE_final
The final kinetic energy can be calculated by equating it to the work done by the electric field:
KE_final = qV
Substituting the values, we get:
KE_final = (2e)(1000 V)
Now, we can calculate the final velocity (v) of the helium ion:
KE_final = (1/2)mv^2
where m is the mass of the helium ion.
Substituting the values, we get:
(2e)(1000 V) = (1/2)m(v^2)
Simplifying, we find:
v = √((4eV)/m)
Now that we have the final velocity (v), we can calculate the radius of curvature (r) using the equation for the centripetal force:
F = (mv^2)/r
In this case, the force is provided by the magnetic field. So, substituting the values, we get:
|q|(v)(B) = (mv^2)/r
Substituting the value of v obtained earlier, we have:
|2e|√((4eV)/m)(B) = (m(4eV)/m)/r
Here, the mass of the helium ion (m) cancels out, leaving us with:
|2e|√((4eV)/m)(B) = (4eV)/r
Simplifying further:
|r| = (2√((4eV)/m)(B))/(4eV)
|r| = (√(4eVB))/(2eV)
|r| = (√(4eB))/(2e)
|r| = (√(4B))/(2)
|r| = √(B)/2
Therefore, the radius of curvature (r) is given by:
r = √(B)/2
Substituting the value of B (1.0 T), we get:
r = √(1.0 T)/2
r = √(1.0) m
r = 1.0 m
Hence, the radius of curvature of the path followed by the helium ion in the magnetic field is 1.0 meter.