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 $\vec { B }$ field.What is the radius of the proton's resulting orbit? (e = 1.60 × 10^-19 C,m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

The radius of the proton's orbit can be calculated using the formula:

r = mv/eB

where 
m = mass of proton = 1.67 × 10^-27 kg
v = final velocity of proton
e = charge of proton = 1.60 × 10^-19 C
B = magnetic field strength = 0.040 T

The final velocity of the proton can be calculated using the conservation of energy:

(1/2)mv^2 = qV
v = √(2qV/m)

where 
q = charge of proton = 1.60 × 10^-19 C
V = potential difference = 1.0 kV = 1000 V

Substituting the values, we get:

v = √(2qV/m) = √(2(1.6×10^-19)(1000)/(1.67×10^-27)) = 9.89×10^5 m/s

Substituting the values for v, e, B, and m, we get:

r = mv/eB = (1.67 × 10^-27 kg)(9.89×10^5 m/s)/(1.60 × 10^-19 C)(0.040 T) = 0.11 m

Therefore, the radius of the proton's orbit is 0.11 m, which is given by choice B.

A Proton,starting from Rest,accelerates Through a Potential Difference of 1 B⃗\vec { B }B

A Proton,starting from Rest,accelerates Through a Potential Difference of 1 $\vec { B }$