Question 8.2: A charged particle of mass 2kg and 1C starts at the origin w...

(a)

F=m\frac{du}{dt}=Qu\times B

a=\frac{du}{dt}=\frac{Q}{m}u\times B

Hence

\frac{d}{dt}(u_{x}a_{x}+u_{y}a_{y}+u_{z}a_{z})=\frac{1}{2}\left|\begin{matrix} a_{x} & a_{y} & a_{z} \\ u_{x} & u_{y} & u_{z} \\ 0 & 0 & 10 \end{matrix}\right| =5(u_{y}a_{x}-u_{x}a_{y})

By equating components, we get

\frac{du_{x}}{dt}=5u_{y}                                                    (8.2.1)

\frac{du_{y}}{dt}=-5u_{x}                                                   (8.2.2)

\frac{du_{z}}{dt}=0\rightarrow u_{z}=C_{o}                                            (8.2.3)

We can eliminate u_{x} or u_{y} in eqs. (8.2.1) and (8.2.2) by taking second derivatives of one equation and making use of the other. Thus

\frac{d^{2}u_{x}}{dt^{2}}=5\frac{du_{y}}{dt}=-25u_{x}

or

\frac{d^{2}u_{x}}{dt^{2}}+25u_{x}=0

which is a linear differential equation with solution (see Case 3 of Example 6.5)

u_{x}=C_{1}\cos5t+C_{2}\sin5t                                              (8.2.4)

From eqs. (8.2.1) and (8.2.4)

5u_{y}=\frac{du_{x}}{dt}=-5C_{1}\sin 5t+5C_{2}\cos5t                        (8.2.5)

or

u_{y}=-C_{1}\sin5t+C_{2}\cos5t

We now determine constants C_{o},  C_{1},  and C_{2} using the initial conditions. At t=0,   u=3a_{y}. Hence

u_{x}=0\rightarrow 0=C_{1}\cdot 1+C_{2}\cdot0\rightarrow C_{1}=0

u_{y}=3\rightarrow 3=-C_{1}\cdot 0+C_{2}\cdot1\rightarrow C_{2}=3

u_{z}=0\rightarrow 0=C_{o}

Substituting the values of C_{o},  C_{1}, and C_{2} into eqs. (8.2.3) to (8.2.5) gives

u=(u_{x},u_{y},u_{z})=(3\sin5t,3\cos5t,0)                        (8.2.6)

Hence

u(t=4)=(3\sin20,3\cos20,0)=2.739a_{x}+1.224a_{y}m/s

a=\frac{du}{dt}=(15\cos5t,-15\sin5t,0)

and

a(t=4)=6.121a_{x}-13.694a_{y}m/s^{2}

(b)

F=ma=12.2a_{x}-27.4a_{y}N

or

F=Qu\times B=(1)(2.739a_{x}+1.224a_{y})\times 10a_{z}=12.2a_{x}-27.4a_{y}N

(c)

K.E.=\frac{1}{2}m|u|^{2}=\frac{1}{2}(2)(2.739^{2}+1.224^{2})=9J

u_{x}=\frac{dx}{dt}=3\sin5t\rightarrow x=-\frac{3}{5}\cos5t+b_{1}                               (8.2.7)

u_{y}=\frac{dy}{dt}=3\cos5t\rightarrow y=\frac{3}{5}\sin5t+b_{2}                                 (8.2.8)

u_{z}=\frac{dz}{dt}=0\rightarrow z=b_{3}                                                                           (8.2.9)

where b_{1},  b_{2}, and b_{3} are integration constants. At t=0, (x,y,z)=(0,0,0) and hence

x(t=0)=0\rightarrow 0=-\frac{3}{5}\cdot1+b_{1}\rightarrow b_{1}=0.6

y(t=0)=0\rightarrow 0=\frac{3}{5}\cdot0+b_{2}\rightarrow b_{2}=0

z(t=0)=0\rightarrow 0=b_{3}

Substituting the values of b_{1},  b_{2}, and b_{3} into eqs. (8.2.7) to (8.2.9), we obtain

(x,y,z)=(0.6-0.6\cos5t,0.6\sin5t,0)                               (8.2.10)

At t=4s

(x,y,z)=(0.3552,0.5478,0)

(d) From eq. (8.2.10), we eliminate t by noting that

(x-0.6)^{2}+y^{2}=(0.6)^{2}(\cos^{2}5t+\sin^{2}5t),      z=0

or

(x-0.6)^{2}+y^{2}=(0.6)^{2},       z=0

which is a circle on plane z=0, centered at (0.6,0,0) and of radius 0.6m. Thus the particle gyrates in an orbit about a magnetic field line.

(e)

K.E.=\frac{1}{2}m|u|^{2}=\frac{1}{2}(2)(9\cos^{2}5t+9\sin^{2}5t)=9J

which is the same as the K.E. at t=0 and t=4s. Thus the uniform magnetic field has no effect on the K.E. of the particle.

Note that the angular velocity \omega =QB/m and the radius of the orbit r=u_{o}/\omega, where u_{o} is the initial speed. An interesting application of the idea in this example is found in a common method of focusing a beam of electrons. The method employs a uniform magnetic field directed parallel to the desired beam as shown in Figure 8.2. Each electron emerging from the electron gun follows a helical path and returns to the axis at the same focal point with other electrons. If the screen of a cathode-ray tube were at this point, a single spot would appear on the screen.