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
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
2 answers
Elena 6.47*10^-3m is wrong answer, please check again. thanks
1 answer