Radius of Curvature of Ion in Magnetic Field

AI Thread Summary
To find the radius of curvature of a doubly ionized atom in a magnetic field, the velocity of the ion must first be calculated using the relationship between kinetic energy and voltage. The conservation of energy principle states that the kinetic energy gained from the voltage (K.E. = qΔV) equals the initial potential energy. The correct formula for the radius of curvature is derived as R = 1/B * (2 * m * ΔV / q)^(1/2). This approach links the concepts of kinetic energy, voltage, and magnetic forces to determine the ion's path in the magnetic field. The discussion emphasizes the importance of correctly applying energy conservation to solve for velocity before calculating the radius.
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Homework Statement



A doubly ionized atom (charge = +2e) whose mass is 5.15E-26 kg is accelerated by a voltage of 3450 V and enters a region where a uniform magnetic field B = 0.100 T acts perpendicular to its motion. What is the radius of curvature of the path of the ion in the B-field?


Homework Equations



r=\frac{mv}{qB}
e = 1.602 x 10-19


The Attempt at a Solution



Given the equation above this is a very simple problem except I don't know how to calculate the velocity of the ion from voltage. The only idea I had was conservation of energy: Kf = Ui or \frac{1}{2}mv2 = q\DeltaV. But this ended up wrong and in retrospect I'm not sure if it entirely makes sense to begin with...
 
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Looks like you are on the right track. The Δ Voltage will result in the ΔPE which becomes KE.

Looks to me like ½mv² gets related to the Lorentz force component and the centripetal acceleration as you've already written.

I end up with an equation that looks like:

R = 1/B*(2 * m * ΔV/ q)1/2
 
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