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?
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))
+5+5
The formula used by Magnets to solve this problem is correct. The formula used is mv^2/R = qvB, where m is the mass of the helium ion, v is its velocity, q is its charge, B is the magnetic field strength, and R is the radius of curvature of the path.
To calculate the radius of curvature, you need to substitute the known values into the formula. The mass of a helium ion (He+) is 4 times the mass of a proton, m = 4(1.67 x 10^-27 kg) = 6.68 x 10^-27 kg. The charge of a helium ion is twice the charge of a proton, q = 2(1.6 x 10^-19 C) = 3.2 x 10^-19 C. The velocity of the helium ion can be derived using the formula v = sqrt(2qΔφ/m), where Δφ is the potential difference. In this case, the potential difference is 1000 V, so v = sqrt(2(3.2 x 10^-19 C)(1000 V)/(6.68 x 10^-27 kg)) ≈ 3.0 x 10^5 m/s. The magnetic field strength B is given as 1.0 T.
Plugging in these values into the formula, we get R = mv/qB = (6.68 x 10^-27 kg)(3.0 x 10^5 m/s) / (3.2 x 10^-19 C)(1.0 T) ≈ 6.62 x 10^-3 m.
So, the radius of curvature of the path of the helium ion is approximately 6.62 x 10^-3 meters.