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IBDP Physics- D.3 Motion in electromagnetic fields- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question 

An electron with speed \(v\) enters the region between two charged parallel plates midway between the plates, as shown. The potential difference between the plates is \(V\).

What is the speed of the electron on impact with the plate?

(A) \(\sqrt{v^2+\dfrac{eV}{2m_{\mathrm{e}}}}\)
(B) \(\sqrt{v^2+\left(\dfrac{eV}{2m_{\mathrm{e}}}\right)^2}\)
(C) \(\sqrt{v^2+\dfrac{eV}{m_{\mathrm{e}}}}\)
(D) \(\sqrt{v^2+\left(\dfrac{eV}{m_{\mathrm{e}}}\right)^2}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The electron starts midway between the plates, so the potential difference through which it moves before reaching the positive plate is \(\dfrac{V}{2}\).

The gain in kinetic energy is equal to the loss in electric potential energy. The magnitude of the charge of an electron is \(e\), so

\(\Delta E_{\mathrm{k}}=e\dfrac{V}{2}\)

Initially, the kinetic energy is

\(E_{\mathrm{k,i}}=\dfrac{1}{2}m_{\mathrm{e}}v^2\)

Therefore, the final kinetic energy is

\(\dfrac{1}{2}m_{\mathrm{e}}v_{\mathrm{f}}^2=\dfrac{1}{2}m_{\mathrm{e}}v^2+\dfrac{eV}{2}\)

Multiplying by \(\dfrac{2}{m_{\mathrm{e}}}\),

\(v_{\mathrm{f}}^2=v^2+\dfrac{eV}{m_{\mathrm{e}}}\)

Hence,

\(v_{\mathrm{f}}=\sqrt{v^2+\dfrac{eV}{m_{\mathrm{e}}}}\)

Therefore, the correct answer is \( \boxed{\mathrm{C}} \).

Question 

An electron moves with speed \(v\) in a region of a uniform magnetic field of strength \(B\). The path of the electron is a circle of radius \(R\). The graph shows how \(R\) varies with \(v\).

What is the gradient of the graph?

(A) \(\frac{m_eB}{e}\)
(B) \(\frac{eB}{m_e}\)
(C) \(\frac{m_e}{eB}\)
(D) \(\frac{e}{m_eB}\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The magnetic force provides the centripetal force for the electron’s circular motion:

\(evB=\frac{m_ev^2}{R}\)

Rearranging for \(R\),

\(R=\frac{m_ev}{eB}\)

This has the form

\(R=\left(\frac{m_e}{eB}\right)v\)

Comparing this with \(y=mx\), the gradient of the graph of \(R\) against \(v\) is

\(\mathrm{gradient}=\frac{m_e}{eB}\)

Thus,

\( \boxed{\mathrm{gradient}=\frac{m_e}{eB}} \)

Hence, the correct answer is \( \boxed{\mathrm{C}} \).

Question

A negatively charged sphere is falling through a magnetic field.
What is the direction of the magnetic force acting on the sphere?
(A) To the left of the page
(B) To the right of the page
(C) Out of the page
(D) Into the page
▶️ Answer / Explanation
Detailed solution

A moving charged particle in a magnetic field experiences a magnetic force given by:

\( \vec{F} = q\,\vec{v} \times \vec{B} \)

The sphere is negatively charged, so the direction of the magnetic force is opposite to the direction given by the right-hand rule for positive charge.

The velocity of the sphere is downward, and the magnetic field is directed from the north pole to the south pole as shown in the diagram.

Using Fleming’s left-hand rule (or reversing the right-hand rule for a negative charge), the magnetic force is directed into the page.

✅ Answer: (D)

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