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

Answer:

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

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

Explanation:

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

$F = qv \times B = qvBsin(\theta)$     (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:

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

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

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

$K = \frac{1}{2}mv^{2}$

Where:

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

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

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

I hope it helps you!