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Magnetic Fields: Question 3

Syllabus 20.3

Structured A2 8 marks

An electron (mass m=9.11×10−31 kgm=9.11\times10^{-31}\text{ kg}, charge of magnitude e=1.60×10−19 Ce=1.60\times10^{-19}\text{ C}) travels through a vacuum at a constant speed of v=2.0×106 m s−1v=2.0\times10^{6}\text{ m s}^{-1}, moving horizontally to the right across the page. It enters a region containing a uniform magnetic field of flux density B=8.0×10−4 TB=8.0\times10^{-4}\text{ T}, directed into the plane of the page, at right angles to its velocity.

(a) Calculate the magnitude of the magnetic force acting on the electron as it enters the field. [2]

(b) State the initial direction of this force on the electron, explaining your reasoning with reference to the electron's negative charge. Explain why this force causes the electron to travel in a circular path at constant speed, and calculate the radius of this path. [4]

(c) Calculate the time taken for the electron to complete one full revolution of its circular path. [2]

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Worked solution

Part (a): Magnitude of the magnetic force

The velocity is at right angles to the field, so θ=90°\theta=90° and sin⁡θ=1\sin\theta=1. Using F=BQvsin⁡θF=BQv\sin\theta with Q=eQ=e: F=8.0×10−4×1.60×10−19×2.0×106×sin⁡90°=2.56×10−16 NF=8.0\times10^{-4}\times1.60\times10^{-19}\times2.0\times10^{6}\times\sin90°=2.56\times10^{-16}\text{ N}

Recomputing by grouping the numbers and powers of ten separately as a check: (8.0×1.60×2.0)=25.6(8.0\times1.60\times2.0)=25.6, and the powers of ten give 10−4−19+6=10−1710^{-4-19+6}=10^{-17}, so F=25.6×10−17=2.56×10−16 NF=25.6\times10^{-17}=2.56\times10^{-16}\text{ N}, the same result.

Part (b): Direction, why the path is circular, and the radius

Direction of the force: for a positive charge moving to the right through a field directed into the page, Fleming’s left-hand rule (first finger = field, into the page; second finger = direction of positive charge flow, to the right) gives a thumb (force) direction vertically upward. The electron, however, carries negative charge, so the force on it is reversed compared with a positive charge moving the same way. Equivalently, the “conventional current” direction to use in Fleming’s rule is to the left (opposite the electron’s actual velocity) for a negative charge moving right. Reapplying the rule with the second finger to the left gives a thumb direction vertically downward. So the electron initially experiences a force directed vertically downward, toward the bottom of the page.

Why the path is circular at constant speed: the magnetic force F=BQvsin⁡θF=BQv\sin\theta always acts perpendicular to the velocity vv (this follows from the vector relation F=Qv×BF=Qv\times B). A force perpendicular to velocity does no work on the electron, since work done =Fdcos⁡90°=0=Fd\cos90°=0, so the force cannot change the electron’s speed. It only ever changes the direction of its velocity. A force of constant magnitude that stays perpendicular to velocity is exactly the condition for uniform circular motion, with the magnetic force providing the centripetal force.

Radius: setting the magnetic force equal to the centripetal force requirement mv2/rmv^2/r: BQv=mv2r  ⟹  r=mvBQBQv=\frac{mv^2}{r}\implies r=\frac{mv}{BQ} r=9.11×10−31×2.0×1068.0×10−4×1.60×10−19=1.822×10−241.28×10−22=1.42×10−2 mr=\frac{9.11\times10^{-31}\times2.0\times10^{6}}{8.0\times10^{-4}\times1.60\times10^{-19}}=\frac{1.822\times10^{-24}}{1.28\times10^{-22}}=1.42\times10^{-2}\text{ m}

Recomputing independently using r=mv2/Fr=mv^2/F with the force from part (a): r=9.11×10−31×(2.0×106)22.56×10−16=3.644×10−182.56×10−16=1.42×10−2 mr=\dfrac{9.11\times10^{-31}\times(2.0\times10^{6})^2}{2.56\times10^{-16}}=\dfrac{3.644\times10^{-18}}{2.56\times10^{-16}}=1.42\times10^{-2}\text{ m}, the same radius both ways, so r≈1.4×10−2 mr\approx1.4\times10^{-2}\text{ m} (1.4 cm) to 2 s.f.

Part (c): Period of the circular motion

The period is independent of speed and can be found directly from T=2πmBQT=\dfrac{2\pi m}{BQ}: T=2π×9.11×10−318.0×10−4×1.60×10−19=2π×9.11×10−311.28×10−22=4.47×10−8 sT=\frac{2\pi\times9.11\times10^{-31}}{8.0\times10^{-4}\times1.60\times10^{-19}}=\frac{2\pi\times9.11\times10^{-31}}{1.28\times10^{-22}}=4.47\times10^{-8}\text{ s}

Recomputing as a check using T=2πr/vT=2\pi r/v with the radius from part (b): T=2π×1.42×10−22.0×106=4.47×10−8 sT=\dfrac{2\pi\times1.42\times10^{-2}}{2.0\times10^{6}}=4.47\times10^{-8}\text{ s}, the same answer both ways, confirming T≈4.5×10−8 sT\approx4.5\times10^{-8}\text{ s} to 2 s.f.

Final answers

  • (a) F=2.56×10−16 NF=\boxed{2.56\times10^{-16}}\text{ N}
  • (b) Force initially downward (toward the bottom of the page); circular motion because the force stays perpendicular to velocity, changing direction only, not speed; radius r=1.42×10−2 mr=\boxed{1.42\times10^{-2}}\text{ m}
  • (c) T=4.47×10−8 sT=\boxed{4.47\times10^{-8}}\text{ s}