IB Physics SL D.3.3 Crossed Electric and Magnetic Fields

Practise balancing electric and magnetic forces in perpendicular fields, selecting undeflected particle speed and predicting how charge sign, mass or field changes alter…

Exam points

draw qE and qvB directions for the stated charge before deciding whether the forces oppose
set qE = qvB to obtain v = E/B for an undeflected particle beam through the selector
predict deflection above or below the selected speed and reverse the pattern for opposite charge

[Maximum number: 2]

Two oppositely charged parallel plates are a distance 8.0 cm apart. The potential difference between the plates is 120 V . An alpha particle is placed on the positively charged plate and released from rest. Gravity is ignored.

Figure for Question D.3.3—Crossed electric and magnetic fields question 1 — IB Physics SL

A magnetic field directed into the plane of the page is now established between the plates. An alpha particle enters the region between the plates with a horizontal speed of $5.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}$ . The particle is not deflected.

Calculate the magnitude of the magnetic field.

$\begin{aligned} F_{\mathrm{m}} =F_{\mathrm{e}} \text { OR } q v B=q E \text { OR } B=F_{\mathrm{e}} / q v \quad B =\ll \frac{E}{v}=\frac{1.5 \times 10^{3}}{5.0 \times 10^{5}}=>3.0 \times 10^{-3} T \end{aligned}$

Award [2] if $3.0 \times 10^{-3}$ « T » is seen as the answer without working.

Allow ECF from a) and b) i)

[2]

All question bank results loaded

D.3.3—Crossed electric and magnetic fields question 1