Question

A beam of protons is accelerated through a potential difference of 0.750kV and then enters a uniform magnetic field traveling perpendicular to the field.

a.What magnitude of field is needed to bend these protons in a circular arc of diameter 1.74m ?

b.What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

Answer 1

If the protons are accelerated through a potential difference then they will gain an energy equivalent to the product of the potential difference and the charge on the proton.

E_p = q_pV

V = 0.750 kV = 750 V = 750 J/C

q_p = 1.602x10^-19 C

E = 1.602x10^-19 * 750 = 1.201x10^-16 J

The electrical energy gained by these protons transforms to kinetic energy i.e. they gain a velocity v.

The two energies can thus be equated

E = m_pv²/2

where m_p = mass of proton = 1.672x10^-27 kg

v = (2E/m_p)^1/2 = 379054.76 m/s

So the proton enters the magnetic field with a velocity of 3.79x10⁵ m/s.

a. The magnetic field acts perpendicular to the velocity of the protons.

The force on these protons is given by

F = q_p(v X B), where v X B is the cross product of velocity and the magnetic field vector

So the magnitude of force is

|F| = q_p|v||B|sin

where is the angle between the velocity and magnetic field = 90^o

As sin90 = 1,

|F| = q_p|v||B|

Now, if the protons bend in a circular arc, then the radial force on the protons must be equal to the electromagnetic force acting on the protons.

The radial force on the protons is given by

|F| = m_p|v|² / R

where R = radius of arc = Diameter/2 = 1.74/2 = 0.87 m

Equating the radial and electromagnetic forces, we get

q_p|v||B| = m_p|v|² / R

|B| = (m_p|v|) / (Rq_p) = 4.55 x 10^-3 T

Ans. The magnitude of the magnetic field required is 4.55x10^-3 T

b. If electrons enter the field with the same speed and were to produce an arc with the same diameter.

Using the above formula for magnitude of field for electrons instead of protons we have

|B| = (m_e|v|)/(Rq_e)

Mass of electron, m_e = 9.109x10^-31 kg

Charge on electron, q_e = -1.602x10^-19 C

We can ignore the minus sign since the problem does not say anything about the direction of the path of the electron.

So,

|B| = (9.109x10^-31 * 3.79x10⁵) / (0.87 * 1.602x10^-19) = 2.48x10^-6 T

Ans. We require a magnetic field of magnitude 2.48x10^-6 T to produce a path with the same diameter as that for the proton.