Question

In a linear accelerator, protons are accelerated from rest through a potential difference to a speed of approximately 3.1 x 10⁶ meters per second. The resulting proton beam produces a current of 2 x 10^-6 ampere.

a. Determine the potential difference through which the protons were accelerated.

b. If the beam is stopped in a target, determine the amount of thermal energy that is produced in the target in one minute.

The proton beam enters a region of uniform magnetic field B, as shown above, that causes the beam to follow a semicircular path.

c. Determine the magnitude of the field that is required to cause an arc of radius 0.10 meter.

d. What is the direction of the magnetic field relative to the axes shown above on the right?

Answer 1

(a) W = K … Vq = ½ mv² … V = (1.67x10^–27)(3.1x10⁶)² / 2(1.6x10^–19 C) … 50000 V

(b) Method I – The thermal energy produced by a single proton will be equal to the conversion of the kinetic energy into internal energy. The kinetic energy can be found with ½ mv² and the v is the same at the target as it was when it entered the B field.

For a single proton we have ½ mv² = ½ (1.67x10^–27)(3.1x10⁶)² = 8x10^–15 J.

Now we have to find out how many protons hit the target in 1 minute using the current.

I = Q/t … 2x10^–6 Amp = Q / 60 sec … Q = 1.2x10^–4 C total charge.

1.2x10^–4 C / 1.6x10^–19 C/proton → 7.5x10¹⁴ protons.

Now multiply the number of protons by the energy for each one. 7.5 x 10¹⁴ * 8 x 10^–15 = 6 J

Alternate (easier) solution – Find the power of the beam P = IV. Then, W = Pt, directly gives the energy that will be delivered in 1 minute. W = IVt = (2x10^–6)(50000)(60) = 6 J

(c) F_net(C) = mv²/r … qvB = mv² / r … B = mv / qr … B = (1.67x10^–27)(3.1x10⁶) / (1.6x10^–19)(0.1) B=0.32 T

(d) Using the RHR gives B field out of the page in the positive z direction.