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IBDP Physics- D.2 Electric and magnetic fields- IB Style Questions For SL Paper 2 -FA 2025

IBDP Physics SL Paper 2 -FA 2025- All Chapters

Question

An electron enters a region of a uniform electric field between two parallel charged plates. The electron is initially halfway between the plates and its initial velocity is parallel to the plates.
The following data are given:
Initial speed of the electron \(= 9.4\times10^{6}\,m\,s^{-1}\)
Potential difference between the plates \(= 30\,V\)
Distance between the plates \(= 4.0\,cm\)
(a) (i) State the direction of the acceleration of the electron.
(ii) Show that the magnitude of the acceleration of the electron is about \(10^{14}\,m\,s^{-2}\).
(b) The electron collides with one of the plates. Determine the distance the electron travels parallel to the plates.
In another experiment, a uniform magnetic field directed into the plane of the diagram is established between the charged plates. The initial velocity of the electron, the distance between the plates and their electric potential difference remain unchanged. The electron passes undeflected through the region of the electric and magnetic fields.

(c) Calculate the magnetic field strength.

Most-appropriate topic codes (IB Physics):

• Topic D.2: Electric and magnetic fields (Motion of charge) — part (a), (b)
• Topic D.3: Motion in electromagnetic fields (Crossed fields) — part (c)
▶️ Answer/Explanation
Detailed solution

(a)
(i) Upwards (towards the positive plate, as the electron is negatively charged).
(ii) Electric field strength \(E = \frac{V}{d} = \frac{30}{0.04} = 750\,V\,m^{-1}\).
Force \(F = qE = 1.60 \times 10^{-19} \times 750 = 1.2 \times 10^{-16}\,N\).
Acceleration \(a = \frac{F}{m} = \frac{1.2 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 1.32 \times 10^{14}\,m\,s^{-2}\).

(b)
Vertical displacement to hit the plate \(s_y = 2.0\,cm = 0.02\,m\) (half the distance).
Using \(s_y = u_y t + \frac{1}{2} a t^2\) with \(u_y = 0\):
\(0.02 = \frac{1}{2} (1.32 \times 10^{14}) t^2 \implies t^2 \approx 3.03 \times 10^{-16} \implies t \approx 1.74 \times 10^{-8}\,s\).
Horizontal distance \(x = v_x t = 9.4 \times 10^6 \times 1.74 \times 10^{-8} \approx 0.16\,m\).

(c)
For undeflected motion, electric force equals magnetic force (\(F_e = F_m\)).
\(qE = qvB \implies B = \frac{E}{v}\).
\(B = \frac{750}{9.4 \times 10^6} \approx 8.0 \times 10^{-5}\,T\).

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