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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/8/b/solution

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Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 8 / b / Solution by

Codex  0 2026-10-25

qE d

on the initially stationary particle. Thus the entry speed into the magnetic region is

v_{0} = \frac{2 qE d}{m} .

(1)

In the source geometry the magnetic field is perpendicular to the page, and the particle enters perpendicular to the straight interface. Its return to that interface occurs after a semicircle. The separation between exit and return points is its orbit diameter, so the required separation

d

gives

r = d /2

. Consequently

∣ B ∣ = \frac{2 m v _{0}}{q d} = \frac{8 m E}{q d} .

(2)

For the drawn field into the page and a positive charge entering upwards, the trajectory bends left. Its speed is constant while in the magnetic field, so the time there is the semicircle length divided by

v_{0}

t_{B} = \frac{π r}{v _{0}} = \frac{π d}{2 v _{0}} = π \frac{m d}{8 qE} = \frac{πm}{q ∣ B ∣} .

(3)

The last form is half the cyclotron period and independently checks the geometric calculation.

Figure 1.
Finite magnetic-strip transmission and a semicircular return to the electric region
. **Details**.

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