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

Question

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
 
I looked at the formula of Bobpursely, for Carl's question, I didn't get it, can you please plug in numberss to see where I have gone wrong. thanks
 
•Magnets - Elena, Saturday, November 23, 2013 at 3:47pm 
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

I calculated the same answer but somehow its wrong, I wonder why? Is there another formula to solve this?

harsha · Answer

i think... work done=0.5*m*v^2 f*d=0.5*m*v^2 e*q*d=0.5*m*v^2 (v/d)*q*d=0.5*m*v^2 velocity=sqrt((2*100*3*e)/(2*Mn+2*Mp)) r=((2*Mn+2*Mp)*velocity)/(b*3*e))

lkug · Answer

+5+5