Radius of Curvature of Ion in Magnetic Field

[Nexest]

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: K_f = U_i or $$\frac{1}{2}$$mv² = q$$\Delta$$V. But this ended up wrong and in retrospect I'm not sure if it entirely makes sense to begin with...

 

[LowlyPion]

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

 

[nlightner]

As a scientist, it is important to always approach problems with accuracy and precision. In this case, the equation r = mv/qB is applicable for a charged particle moving in a uniform magnetic field. However, in order to calculate the radius of curvature, we need to know the velocity of the ion. In this scenario, we can use the conservation of energy principle to solve for the velocity.

The initial kinetic energy (K_i) of the ion can be calculated using the given voltage and the charge of the ion (q = +2e). K_i = q\DeltaV = (2)(1.602 x 10^-19)(3450) = 1.108 x 10^-15 J.

We can then equate this to the final kinetic energy (K_f) of the ion, which can be calculated using the mass and velocity of the ion. K_f = \frac{1}{2}mv^2.

Setting these equal to each other, we get 1.108 x 10^-15 J = \frac{1}{2}(5.15 x 10^-26)(v^2). Solving for v, we get v = 4.43 x 10^6 m/s.

Now, we can plug this value into the equation r = mv/qB to calculate the radius of curvature. r = \frac{(5.15 x 10^-26)(4.43 x 10^6)}{(2)(1.602 x 10^-19)(0.100)} = 0.00135 m.

Therefore, the radius of curvature of the path of the ion in the magnetic field is 0.00135 m. It is important to note that this is a simplified calculation and does not take into account factors such as the ion's initial direction and the shape of the magnetic field. Further analysis and experimentation may be necessary for a more accurate calculation.