Question

A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040 T at a right angle to the field. What is the radius of the proton's resulting orbit? (e = 1.60 × 10-19 C, mproton = 1.67 × 10-27 kg)

Answer 1

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ($F_{B}$), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

Where:

m: is the proton's mass =  1.67*10⁻²⁷ kg

v: is the proton's velocity

r: is the radius of the proton's orbit

q: is the proton charge = 1.6*10⁻¹⁹ C

B: is the magnetic field = 0.040 T

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$   (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$   (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

ΔV: is the potential difference = 1.0 kV

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!