Determine the direction of the magnetic force on a beam of charged particles or on a current‑carrying conductor when it moves through a magnetic field, and describe the classic experiment that demonstrates this force (including the effect of reversing the current and the magnetic field).
\[
\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}
\]
\[
\mathbf{F}=I\,\mathbf{L}\times\mathbf{B}
\]
\(\mathbf{L}\) points in the direction of conventional current (positive to negative).
\[
F = I L B \sin\theta
\]
\(\theta\) = angle between \(\mathbf{L}\) (or \(\mathbf{v}\)) and \(\mathbf{B}\).
Apparatus
Procedure
Observations
Practical tip – measuring the force
\[
F = \frac{m g d_{\text{cm}} \sin\alpha}{d}
\]
(with \(m\) the mass of the wire segment and \(g = 9.8\ \text{m s}^{-2}\)).
| Case | Current \(\mathbf{I}\) | Magnetic Field \(\mathbf{B}\) | Resulting Force \(\mathbf{F}\) (thumb direction) |
|---|---|---|---|
| (i) | Original direction | Original direction | Given by the right‑hand rule |
| (ii) | Reversed (\(\mathbf{I}\rightarrow -\mathbf{I}\)) | Original | Force reverses (opposite to case i) |
| (iii) | Original | Reversed (\(\mathbf{B}\rightarrow -\mathbf{B}\)) | Force reverses (opposite to case i) |
| (iv) | Reversed | Reversed | Force returns to the direction of case i (two reversals cancel) |
Problem: A beam of protons (\(q=+e\)) travels eastward with speed \(v\) and enters a uniform magnetic field that points vertically upward. What is the direction of the magnetic force on the protons?
Solution
If the same beam were made of electrons (\(q=-e\)), the force would be opposite to the thumb direction. Using the right‑hand rule as above gives a thumb pointing south; the actual force on the electrons would therefore be northward.
Problem: A horizontal wire carries a current of 5 A from left to right. It is placed in a uniform magnetic field of 0.2 T that points into the page. Determine the direction of the magnetic force on the wire and calculate its magnitude if the wire segment inside the field is 0.10 m long.
Solution
\[
F = I L B \sin 90^{\circ}= (5\ \text{A})(0.10\ \text{m})(0.2\ \text{T}) = 0.10\ \text{N}.
\]
If the velocity (or current) is parallel or antiparallel to the magnetic field (\(\mathbf{v}\parallel\mathbf{B}\) or \(\mathbf{I}\parallel\mathbf{B}\)), then \(\theta = 0^{\circ}\) or \(180^{\circ}\) and
\[
\sin\theta = 0\;\;\Longrightarrow\;\;\mathbf{F}=0.
\]
The charge (or conductor) moves straight through the field without any deflection.
The magnetic force on a moving charge or on a current‑carrying conductor is given by the cross‑product \(\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\) (or \(\mathbf{F}=I\,\mathbf{L}\times\mathbf{B}\)). Its magnitude is \(F = I L B \sin\theta\). The direction is found with the right‑hand rule for positive charges (or conventional current) and reversed for negative charges. The horseshoe‑magnet experiment provides a clear, observable demonstration of the force and shows how reversing the current or the magnetic field reverses the force direction. Mastery of these concepts enables students to predict the behaviour of particle beams and conductors in magnetic fields – a core requirement of the Cambridge IGCSE Physics (0625) syllabus.
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