A proton enters, at the origin, a…

A proton enters, at the origin, a region having uniform magnetic field B=0.20 T along +z and uniform electric field E=1.0×10^4 V m^-1 also along +z.…

Numerical answer

A proton enters, at the origin, a region having uniform magnetic field

B=0.20\ \mathrm{T}

along

+z

and uniform electric field

E=1.0\times10^{4}\ \mathrm{V\,m^{-1}}

also along

+z

. Its initial speed is

2.0\times10^{6}\ \mathrm{m\,s^{-1}}

60^\circ

to the

z

-axis. Use

m_p=1.67\times10^{-27}\ \mathrm{kg}

and

e=1.6\times10^{-19}\ \mathrm{C}

. The distance (in cm) moved along the

z

-axis by the proton during its first complete revolution about the

z

-axis (one cyclotron period) is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 37.94

Because

\vec E\parallel\vec B

, the motion splits in two. In the

xy

-plane the proton circles with the cyclotron period

T=\dfrac{2\pi m}{eB}=\dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.20)}=3.279\times10^{-7}\ \mathrm{s}

, unaffected by

E

because the electric force has no component in the plane. Along

z

the proton has initial speed

v_\parallel=v\cos60^\circ=1.0\times10^{6}\ \mathrm{m\,s^{-1}}

and constant acceleration

a=eE/m=\dfrac{1.6\times10^{-19}\times10^{4}}{1.67\times10^{-27}}=9.581\times10^{11}\ \mathrm{m\,s^{-2}}

. In one period,

z=v_\parallel T+\tfrac12aT^2=0.3279+0.0515=0.3794\ \mathrm{m}=37.94\ \mathrm{cm}

. The first term alone,

32.79

cm, would be the pitch without the electric field.

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