Question

A proton traveling at 25.5° with respect to the direction of a magnetic field of strength 2.07 mT experiences a magnetic force of 5.12 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

Answer 1

Answer 2

To calculate the proton's speed and kinetic energy in electron-volts (eV), we'll use the following equations:

(a) The magnetic force on a charged particle moving through a magnetic field is given by the formula:

F = q * v * B * sin(θ)

where: F is the magnetic force on the proton (given as 5.12 × 10⁻¹⁷ N), q is the charge of the proton (1.6 × 10⁻¹⁹ C), v is the speed of the proton (what we need to find), B is the magnetic field strength (given as 2.07 mT, which is 2.07 × 10⁻³ T), θ is the angle between the velocity of the proton and the magnetic field (given as 25.5°).

(b) The kinetic energy (KE) of the proton can be calculated using the formula:

KE = (1/2) * m * v²

where: m is the mass of the proton (1.67 × 10⁻²⁷ kg).

Let's proceed with the calculations:

(a) Calculate the proton's speed (v):

We rearrange the magnetic force equation to solve for v: v = F / (q * B * sin(θ))

v = 5.12 × 10⁻¹⁷ N / (1.6 × 10⁻¹⁹ C * 2.07 × 10⁻³ T * sin(25.5°))

Using trigonometric identities, sin(25.5°) ≈ 0.438371, so:

v ≈ 5.12 × 10⁻¹⁷ N / (1.6 × 10⁻¹⁹ C * 2.07 × 10⁻³ T * 0.438371) v ≈ 5.12 × 10⁻¹⁷ N / (5.22 × 10⁻²⁵ C·T)

Now, we'll convert the magnetic field strength to SI units (T = kg·C⁻¹·s⁻¹):

B = 2.07 × 10⁻³ T = 2.07 × 10⁻³ kg·C⁻¹·s⁻¹

v ≈ 5.12 × 10⁻¹⁷ N / (5.22 × 10⁻²⁵ C·T) ≈ 9.81 × 10⁷ m/s

(b) Calculate the kinetic energy (KE) in electron-volts (eV):

KE = (1/2) * m * v²

KE = (1/2) * (1.67 × 10⁻²⁷ kg) * (9.81 × 10⁷ m/s)²

KE ≈ 8.18 × 10⁻¹⁴ J

To convert the energy from joules to electron-volts (eV), we'll use the conversion factor:

1 eV ≈ 1.6 × 10⁻¹⁹ J

KE in eV ≈ (8.18 × 10⁻¹⁴ J) / (1.6 × 10⁻¹⁹ J/eV) ≈ 5.11 × 10⁵ eV

So, the proton's speed is approximately 9.81 × 10⁷ m/s, and its kinetic energy is approximately 5.11 × 10⁵ eV.