 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/1/iii/solution

Let

r_{P}

be the initial position,

\hat{b} = B / B

, and

Ω = qB / m

, including the sign of

q

. Solving

\dot{v} = Ω v \times \hat{b}

gives

v (t) = v_{0} cos Ω t + (v_{0} \times \hat{b}) sin Ω t .

Thus

r (t) = r_{P} + \frac{v _{0}}{Ω} sin Ω t + \frac{v _{0} \times b ^}{Ω} (1 - cos Ω t) .

The locus is a circle in the plane perpendicular to

B

, centred at

r_{c} = r_{P} + (v_{0} \times \hat{b}) /Ω

, with

∣ r - r_{c} ∣ = \frac{m ∣ v _{0} ∣}{∣ q ∣ B}, (r - r_{P}) \cdot B = 0.

This also follows by equating the magnetic force to the centripetal force. The nonrelativistic angular frequency is

∣ q ∣ B / m

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