Question

A beam of protons is accelerated through a potential difference of 0.720 kV 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 1.79 m ?

Express your answer in tesla to three significant figures.

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?

Express your answer in tesla to three significant figures.

Answer 1

A)
the radius of this curve is

=1.79/2

=0.895 m
Here's the equation for the radius of curvature of a charge through a magnetic field:
r = (mv)/(Bq)
radius, r
mass, m
velocity, v
magnetic field, B
charge, q

The radius is .895 m
The mass of a proton is 1.67x10^-27 kg
The charge of a proton is 1.6x10^-19 C
The only other thing we need to know to solve for B is the velocity.
V = J/Q
The voltage is 720 V and we know the charge of a proton from above so we can solve for the energy the proton gained by going through that potential difference.
J = 1.152x10^-16 J
All this energy would be kinetic energy for the proton so
1.152x10^-16 J = 1/2mv²
We know mass so you can solve for velocity, v
v = 379473.3192 m/s

Now we have everything we need to know to solve for B
B =mv/qr

B=1.6*10^-27*379473.3192/1.6*10^-19*0.895

B=4.24*10^-3 T

B)

Now only the mass of the particle will change so,

B=mv/qr

B=9.1*10^-31*379473.3192/1.6*10^-19*0.895

B=2.41*10^-6 T