A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path?
A) 15 mm
B) 12 mm
C) 18 mm
D) 24 mm
E) 8.5 mm

Question

Quizplus · Accepted Answer

The speed of the proton after being accelerated through the potential difference can be found using the formula:

ΔV = KE/q

where ΔV is the potential difference, KE is the kinetic energy, and q is the charge of the particle.

Solving for KE, we get:

KE = (1/2)qΔV

Plugging in the values, we get:

KE = (1/2)(1.602 x 10^-19 C)(2.5 x 10^3 V) = 2.003 x 10^-16 J

We can then use the formula for the radius of the path of a charged particle moving perpendicularly through a magnetic field:

r = mv/qB

where m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field strength.

Since the proton is moving perpendicularly through the magnetic field, its velocity will be given by:

v = (2KE/m)^0.5

where m is the mass of the proton.

Plugging in the values, we get:

v = (2(2.003 x 10^-16 J)/(1.673 x 10^-27 kg))^0.5 = 5.599 x 10^6 m/s

Finally, plugging in all the values, we get:

r = (1.673 x 10^-27 kg)(5.599 x 10^6 m/s)/(1.602 x 10^-19 C)(0.60 T) = 1.202 x 10^-2 m = 12 mm

So the answer is B, with a radius of 12 mm.

A Proton Is Accelerated from Rest Through a Potential Difference