Question

A proton traveling at 30.9∘with respect to the direction of a magnetic field of strength 3.16mT experiences a magnetic force of 8.57× 10⁻¹⁷N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts. (a) Number Units (b) Number Units

Answer 1

(a) The proton's speed is approximately 2.00 × 10^6 m/s.

(b) Its kinetic energy is roughly 8.05 × 10^-13 eV.

Step-by-step explanation:

(a) To calculate the proton's speed, we utilize the formula for the magnetic force (\(F\)) acting on a charged particle moving through a magnetic field: \(F = qvB\sin(\theta)\), where \(q\) is the charge of the particle, \(v\) is its velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity vector and the magnetic field direction.

Solving for \(v\), we get
`\(v = (F)/(qB\sin(\theta))\). Given \(F = 8.57 * 10^(-17)\) N, \(B = 3.16 * 10^(-3)\) T, and \(\theta = 30.9^\circ\), and considering the charge of a proton (\(q = 1.6 * 10^(-19)\) C),` we can compute \(v\) which equals approximately \(2.00 \times 10^6\) m/s.

`(b) The kinetic energy (\(KE\)) of the proton can be found using the formula \(KE = (1)/(2)mv^2\), where \(m\) is the mass of the proton. With the speed calculated in (a) and the mass of a proton (\(m = 1.67 * 10^(-27)\) kg), the kinetic energy can be determined.`

However, to convert kinetic energy to electron-volts (eV), one can utilize the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19}\) J. After conversion, the kinetic energy of the proton is approximately
`\(8.05 * 10^(-13)\) eV.`

Understanding the relationship between magnetic forces, velocities, kinetic energy, and charge on particles provides insight into particle behavior in magnetic fields and contributes to various applications in physics, such as particle accelerators and magnetic confinement in fusion reactors.