Question

A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

Answer 1

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.

Step-by-step explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

`F = qv * B = qvBsin(\theta)` (1)

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

`v = (F)/(qBsin(\theta)) = (9.14 \cdot 10^(-17) N)/(1.602\cdot 10^(-19) C*3.28 \cdot 10^(-3) T*sin(17.6)) = 5.75 \cdot 10^(5) m/s`

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

`K = (1)/(2)mv^(2)`

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

`K = (1)/(2)mv^(2) = (1)/(2)1.67 \cdot 10^(-27) kg*(5.75 \cdot 10^(5) m/s)^(2) = 2.76 \cdot 10^(-16) J*(1 eV)/(1.602 \cdot 10^(-19) J) = 1723 eV`

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!