Paper 4, Section II, C

Dynamics and Relativity | Part IA, 2015

A particle of mass mm and charge qq has position vector r(t)\mathbf{r}(t) and moves in a constant, uniform magnetic field B\mathbf{B} so that its equation of motion is

mr¨=qr˙×Bm \ddot{\mathbf{r}}=q \dot{\mathbf{r}} \times \mathbf{B}

Let L=mr×r˙\mathbf{L}=m \mathbf{r} \times \dot{\mathbf{r}} be the particle's angular momentum. Show that

LB+12qr×B2\mathbf{L} \cdot \mathbf{B}+\frac{1}{2} q|\mathbf{r} \times \mathbf{B}|^{2}

is a constant of the motion. Explain why the kinetic energy TT is also constant, and show that it may be written in the form

T=12mu((uv)vr2u¨)T=\frac{1}{2} m \mathbf{u} \cdot\left((\mathbf{u} \cdot \mathbf{v}) \mathbf{v}-r^{2} \ddot{\mathbf{u}}\right)

where v=r˙,r=r\mathbf{v}=\dot{\mathbf{r}}, r=|\mathbf{r}| and u=r/r\mathbf{u}=\mathbf{r} / r.

[Hint: Consider u u˙.]\cdot \dot{\mathbf{u}} .]

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