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9702 · 20.3

Force on a moving charge — practice questions

Practice and worked examples for 9702 Force on a moving charge. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A proton enters a velocity selector with an electric field strength of 2.0×104 V m−12.0 \times 10^4 \, \text{V m}^{-1} and a magnetic flux density of 0.50 T0.50 \, \text{T}. Calculate the speed at which the proton will pass through undeflected.

Show solution outline
  1. Identify the given values: E=2.0×104 V m−1E = 2.0 \times 10^4 \, \text{V m}^{-1}, B=0.50 TB = 0.50 \, \text{T}.
  2. Recall the formula for selected velocity in a velocity selector: v=EBv = \frac{E}{B}.
  3. Substitute the values into the formula: v=2.0×104 V m−10.50 Tv = \frac{2.0 \times 10^4 \, \text{V m}^{-1}}{0.50 \, \text{T}}.
  4. Calculate the speed: v=4.0×104 m s−1v = 4.0 \times 10^4 \, \text{m s}^{-1}.
  5. State the answer with units: The proton will pass through undeflected at a speed of 4.0×104 m s−14.0 \times 10^4 \, \text{m s}^{-1}.

Worked example 2

An electron travels at a speed of 5.0 x 10^6 m/s and enters a region of uniform magnetic field of flux density 0.020 T. The electron's path is perpendicular to the magnetic field. Calculate the radius of the circular path it follows. (Charge of an electron = 1.60 x 10^-19 C, mass of an electron = 9.11 x 10^-31 kg)

Show solution outline
  1. Identify the forces acting on the electron. The magnetic force provides the centripetal force for the circular motion.
  2. Equate the formula for magnetic force (FB=BQvF_B = BQv) and centripetal force (Fc=mv2rF_c = \frac{mv^2}{r}), since θ=90∘\theta = 90^\circ and sin⁡(90∘)=1\sin(90^\circ)=1. BQv=mv2rBQv = \frac{mv^2}{r}
  3. Rearrange the equation to solve for the radius, rr. r=mv2BQv=mvBQr = \frac{mv^2}{BQv} = \frac{mv}{BQ}
  4. Substitute the given values into the equation:
    • m=9.11×10−31 kgm = 9.11 \times 10^{-31} \, \text{kg}
    • v=5.0×106 m s−1v = 5.0 \times 10^6 \, \text{m s}^{-1}
    • B=0.020 TB = 0.020 \, \text{T}
    • Q=1.60×10−19 CQ = 1.60 \times 10^{-19} \, \text{C} r=(9.11×10−31)×(5.0×106)(0.020)×(1.60×10−19)r = \frac{(9.11 \times 10^{-31}) \times (5.0 \times 10^6)}{(0.020) \times (1.60 \times 10^{-19})}
  5. Calculate the result: r=4.555×10−243.2×10−21=1.423×10−3 mr = \frac{4.555 \times 10^{-24}}{3.2 \times 10^{-21}} = 1.423 \times 10^{-3} \, \text{m}
  6. State the final answer with appropriate significant figures and units: The radius of the circular path is 1.4×10−3 m1.4 \times 10^{-3} \, \text{m} or 1.4 mm1.4 \, \text{mm}.
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