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An alpha particle travels in a circular path of radius $0.45$m in a magnetic field with $B=1.2$ w/m$^2$. Calculate (i) its speed (ii) its period of revolution, and (iii) its kinetic energy. Mass of proton particle = \(1.67\times 10^{-27}\)kg \(\approx 4\times M_p= 4\times938.27\) MeV.

Solution :

• [(i)] the magnetic force \(eBv\) must be equal to the mass times acceleration. Therefore \begin{equation*} Bev = \frac{Mv^2}{R}, \end{equation*} where \(R\) is the radius of the circular orbit. Hence \begin{equation*} v= \frac{eBR}{M} = \frac{2\times1.6 \times10^{-19}\times 1.2 \times 0.45}{4\times 1.67\times 10^{-27}}\approx 2.7 \times10^7 \text{m/s}. \end{equation*}
• [(ii)] The time period is \begin{equation*} T = \frac{2\pi R}{v} = \frac{2\times3.14\times 0.45}{2.7\times10^7} \approx10^{-7} \text{ s}. \end{equation*}
• [(iii)] The kinetic energy is given by \begin{eqnarray}\nonumber \text{K.E.} &=& \frac{1}{2} M v^2= \frac{1}{2}\times (4\times 1.67 \times 10^{-27}) \times \big(2.7\times10^7\big)^2 \\\nonumber &=& 3.26\times 7.29 \times 10^{-13} \approx 23.7 \times 10^{-13} \text{J}. \end{eqnarray}

A "dipole" is formed from a rod of length \(2a\) and two charges \(+q\)
and \(-q\). Two such dipoles are oriented as shown in figure at the end,
their centers being separated by a distance \(R\). Calculate the force
exerted on the left dipole and show that, for \(R>>a\), the force is
approximately given by
\[F=\frac{3p^2}{2\pi\epsilon_0R^4}\]
where \(p=2qa\) is the dipole moment.

An electron moving with speed of $5.0\times 10^8$cm/sec is shot parallel to an electric field strength of $1.0\times 10^3 $nt/coul arranged so as to retard its motion.

• How far will the electron travel in the field before coming (momentarily) to rest ?
• how much time will elapse?
• If the electric field ends abruptly after $0.8$ cm, what fraction of its initial energy will the electron loose in traversing the field?

Prove that the potential \(\phi(0)\) at the center of a charge-free spherical volume \(V\) is the equal to average of \(\phi(r)\) over the surface \(S\) of the sphere. Using 

1. Green’s reciprocity relation.
2. Green's identity\[\iiint d^3r (f \nabla^2 g − g\nabla^2 f ) = \iint \hat{n} · (f \nabla g − g\nabla f) dS\]

Six charges are placed at positions given below:

Compute the dipole moment of the system.

The permanent dipole moment of water molecule is given to be \(1.86\times10^{-18}\) esc-cm.  How much work is done on water molecule if electric field is increased from  zero to 100 volts/cm and the permanent dipoe is rotated from \(\theta=0\) to \(\theta=\pi\)? 
(Be careful about units.)

Atlee Jackson