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
mv²/2=qΔφ
v=sqrt(2qΔφ/m) =
=sqrt(2•2•1.6•10⁻¹⁹•1000/4•1.67•10⁻²⁷) = 3.1•10⁵ m/s
mv²/R = qvB
R=mv/qB = 4•1.67•10⁻²⁷•3.1•10⁵/2•1.6•10⁻¹⁹•1=
=6.47 •10⁻³m
Elena 6.47*10^-3m is wrong answer, please check again. thanks
To find the radius of curvature of the path taken by the helium ion in the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field.
The formula for the magnetic force is given by:
F = qvB sin(θ)
Where:
- F is the magnitude of the magnetic force
- q is the charge of the particle (in this case, the charge of the helium ion, which is equal to the charge of an electron, -1.6 x 10^(-19) C)
- v is the velocity of the particle
- B is the magnetic field strength (given as 1.0 T in this case)
- θ is the angle between the velocity vector of the particle and the magnetic field vector
The magnetic force provides the necessary centripetal force to keep the particle moving in a circular path. The centripetal force is given by:
F = (mv^2) / r
Where:
- m is the mass of the particle (in this case, the mass of the helium ion, which is approximately 6.64 x 10^(-27) kg)
- v is the velocity of the particle
- r is the radius of curvature of the path
By equating the magnetic force and the centripetal force, we can derive an expression for the radius of curvature:
qvB sin(θ) = (mv^2) / r
Since the helium ion is released from rest, its initial velocity (v) will be zero. Therefore, the angle (θ) between the velocity vector and the magnetic field vector is also zero. This means that sin(θ) = 0, and the equation simplifies to:
qvB = mv^2 / r
Simplifying further:
r = mv / (qB)
Now we can substitute the known values to find the radius of curvature:
m = 6.64 x 10^(-27) kg
v = 0 m/s (released from rest)
q = 1.6 x 10^(-19) C (charge of the helium ion)
B = 1.0 T
r = (6.64 x 10^(-27) kg * 0 m/s) / (1.6 x 10^(-19) C * 1.0 T)
r = 0
The radius of curvature is therefore zero, which means that the path of the helium ion is a straight line. This result might seem counterintuitive, but it occurs because the magnetic force acts perpendicular to the velocity of the particle and does not deflect the ion from its original path.
Please note that all the calculations were based on the given values and assumptions provided in the question.