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?

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin $\mathrm{\theta },\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

Cp= (0, -Rsin $\mathrm{\theta },-\frac{3}{2}\mathrm{R}-\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

The circles of the two

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This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is

Ce= (0, -Rsin $\mathrm{\theta },\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

Cp= (0, -Rsin $\mathrm{\theta },-\frac{3}{2}\mathrm{R}-\mathrm{R}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }$ )

The circles of the two shall not overlap if the distance between the two centers are

Greater than 2R.

Let d be the distance between Cp and Ce.

Let d be the distance between Cp and Ce.

So d²= (2Rsin $\theta$ )²+ ( $\frac{3}{2}R-2Rcos\theta$ )²

= 4R²sin² $\theta$ + $\frac{9}{4}R$ ²-6R²cos $\theta$ +4R²cos² $\theta$

= 4R²+ $\frac{9}{4}R$ ²-6R²cos $\theta$

Since d has to be greater than 2R

So d²> 4R²

4R²+ $\frac{9}{4}R$ ²-6R²cos $\theta$ > 4R²

Cos $\theta$ <3/8

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<p>This is a Long Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation-</strong> Since, B is along the x-axis, for a circular orbit the momenta of the two particles are in the y-z plane. Let P1 and P2 be the momentum of the electron and positron, respectively. Both traverse a circle of radius R of opposite sense. Let P1 make an angle? with the y-axis P2 must make the same angle.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1742967460php6x1nij_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1742967460php6x1nij.jpeg" alt="" width="244" height="243"></picture></div><div><p>The centres of the respective circles must be perpendicular to the momenta and at a distance R. Let the centre of the electron be at Ce and of the positron at Cp . The coordinates of Ce is</p><p>Ce= (0, -Rsin&nbsp;<math><mi mathvariant="normal">θ</mi><mo>, </mo><mi mathvariant="normal">R</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">θ</mi></math>&nbsp;)</p><p>Cp= (0, -Rsin&nbsp;<math><mi mathvariant="normal">θ</mi><mo>, </mo><mo>-</mo><mfrac><mrow><mrow><mn>3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi mathvariant="normal">R</mi><mo>-</mo><mi mathvariant="normal">R</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">θ</mi></math>&nbsp;)</p><p>The circles of the two shall not overlap if the distance between the two centers are</p><p>Greater than 2R.</p><p>Let d be the distance between Cp and Ce.</p><p>Let d be the distance between Cp and Ce.</p><p>So d²= (2Rsin <math><mi>θ</mi></math> )²+ (&nbsp;<math><mfrac><mrow><mrow><mn>3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi>R</mi><mo>-</mo><mn>2</mn><mi>R</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi></math> )²</p><p>= 4R²sin² <math><mi>θ</mi></math>&nbsp;+&nbsp;<math><mfrac><mrow><mrow><mn>9</mn></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac><mi>R</mi></math> ²-6R²cos&nbsp;<math><mi>θ</mi></math> +4R²cos² <math><mi>θ</mi></math></p><p>= 4R²+ <math><mfrac><mrow><mrow><mn>9</mn></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac><mi>R</mi></math> ²-6R²cos&nbsp;<math><mi>θ</mi></math></p><p>Since d has to be greater than 2R</p><p>So d²&gt; 4R²</p><p>4R²+ <math><mfrac><mrow><mrow><mn>9</mn></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac><mi>R</mi></math> ²-6R²cos&nbsp;<math><mi>θ</mi></math> &gt; 4R²</p><p>Cos&nbsp;<math><mi>θ</mi></math>&nbsp;&lt;3/8</p></div></div>

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