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  • An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R) respectively, in a uniform magnetic field B = B_0i, each with an equal momentum of magnitude p = eBR.

An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R) respectively, in a uniform magnetic field B = B0i, each with an equal momentum of magnitude p = eBR. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Answers (1)

Due to the presence of the magnetic field B along the x-axis, the momenta of two particles in a circular orbit is in the y-z plane. Assume the momenta of the electron (e– ) and positron (e+) to be p1 and p2, respectively. Both move in the opposite sense in a circle of radius R. Let p, make an angle θ with the y-axis p2 must make the same angle with the axis.

But the respective circles must have their centres perpendicular to the momenta at a distance R. Let Ce be the centre of the electron and Cp of the positron.

The coordinates of Ce are Ce = (0, -R \sin \theta, R \cos \theta)

The coordinates of Cp are Cp = [0, -R \sin \theta, (1.5R - R \cos \theta)]

The circular orbits of electron and positron shall not overlap if the distance between the two centres is greater than 2R.

Let d be the distance between Cp and Ce. Then

d^{2}=[R\sin \theta -(-R \sin \theta )]^{2}\left [ R\cos \theta \left ( \frac{3}{2}R-R\cos \theta \right ) \right ]^{2}

=(2R\sin \theta)^{2}+\left ( 2R \cos \theta -\frac{}3{}2 R \right )^{2}

=4R^{2}\sin ^{2}\theta +4R^{2}\cos^{2}\theta -6R^{2}\cos \theta +\frac{9}{4}R^{2}

=4R^{2} -6R^{2}\cos \theta +\frac{9}{4}R^{2}

As d has to be greater than 2R d2 > 4R2

=4R^{2} -6R^{2}\cos \theta +\frac{9}{4}R^{2}>4R^{2}

\frac{9}{4}>6 \cos \theta or \cos \theta <\frac{3}{8}

 

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