Paper 4 Magnetic Force Answers

Q1 Current Balance

1. Axes correctly oriented and labelled [1]; sensible scales [1]; all six points plotted accurately [2]; best-fit straight line [1].

2. The graph gradient is approximately $2.14\mathrm{g}{\mathrm{A}}^{-1}=2.14\times 10^{-3}\mathrm{kg}{\mathrm{A}}^{-1}$. Accept graph-consistent values and require the unit conversion. [2]

3. Since $F=\Delta mg=BIL$,

   $$B=\frac{g}{L}\frac{\Delta m}{\Delta I}=\frac{9.81}{0.0600}(2.14\times 10^{-3})\approx 0.350\mathrm{T}.$$

   Accept a value consistent with the candidate’s gradient. [3]

4. In general $F=BIL\sin \theta$. Perpendicular wire and field give $\theta =90^{\circ }$ and $\sin \theta =1$, so $F=BIL$. [2]

5. Use a low current, switch off between readings, or take readings promptly. [1]

Q2 Beam Deflection Comparison

1. Electric deflection depends on electric-field direction and can change speed/energy; magnetic deflection is perpendicular to velocity and magnetic field and does no work. A crossed-field no-deflection setup can distinguish them. [2]
2. Magnetic force direction follows $q\vec{v}\times \vec{B}$; reversing $\vec{B}$ reverses the vector product. [2]
3. Avoid touching high-voltage terminals, switch off before changing connections, or follow tube voltage limits. [1]