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Magnetic Fields: Question 7

Syllabus 20.3

Structured A2 8 marks

A beam of protons (mass mp=1.67×10−27 kgm_p=1.67\times10^{-27}\text{ kg}, charge +e=1.60×10−19 C+e=1.60\times10^{-19}\text{ C}) enters a velocity selector. Inside the selector, a uniform electric field of strength E=3.6×105 V m−1E=3.6\times10^{5}\text{ V m}^{-1} and a uniform magnetic field of flux density B1=0.45 TB_1=0.45\text{ T} are perpendicular to each other, and both are perpendicular to the initial velocity of the protons, arranged so that the electric force and the magnetic force on a proton act in opposite directions.

(a) Calculate the speed vv of the protons that pass straight through the selector without being deflected. [2]

(b) These protons then leave the selector and enter a separate region containing only a uniform magnetic field of flux density B2=0.20 TB_2=0.20\text{ T}, directed at right angles to their velocity. Calculate the radius of the circular path followed by the protons in this second region. [3]

(c) Calculate the time taken for a proton to complete a quarter of a full revolution of this circular path. [3]

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Worked solution

Part (a): Speed selected by the velocity selector

For a proton to pass straight through undeflected, the electric force and magnetic force on it must balance: QE=QB1vQE=QB_1v

The charge QQ cancels from both sides, leaving: v=EB1=3.6×1050.45=8.0×105 m s−1v=\frac{E}{B_1}=\frac{3.6\times10^{5}}{0.45}=8.0\times10^{5}\text{ m s}^{-1}

Recomputing as a check: 3.6÷0.45=8.03.6\div0.45=8.0, and the powers of ten are unchanged (10510^{5}), so v=8.0×105 m s−1v=8.0\times10^{5}\text{ m s}^{-1}, the same result.

Part (b): Radius of the circular path in the second field

In the second region, the magnetic force provides the centripetal force: B2Qv=mpv2r  ⟹  r=mpvB2QB_2Qv=\frac{m_pv^2}{r}\implies r=\frac{m_pv}{B_2Q}

r=1.67×10−27×8.0×1050.20×1.60×10−19=1.336×10−213.2×10−20=4.175×10−2 mr=\frac{1.67\times10^{-27}\times8.0\times10^{5}}{0.20\times1.60\times10^{-19}}=\frac{1.336\times10^{-21}}{3.2\times10^{-20}}=4.175\times10^{-2}\text{ m}

Recomputing as a check, grouping numbers and powers of ten separately: 1.67×8.0=13.361.67\times8.0=13.36 with powers 10−27+5=10−2210^{-27+5}=10^{-22}, giving a numerator of 13.36×10−22=1.336×10−2113.36\times10^{-22}=1.336\times10^{-21}; the denominator is 0.20×1.60=0.320.20\times1.60=0.32 with powers 10−1910^{-19}, giving 3.2×10−203.2\times10^{-20}. Dividing, 1.336÷3.2=0.41751.336\div3.2=0.4175 with powers 10−21−(−20)=10−110^{-21-(-20)}=10^{-1}, so r=0.4175×10−1=4.175×10−2 mr=0.4175\times10^{-1}=4.175\times10^{-2}\text{ m}, the same result, so r≈4.2×10−2 mr\approx4.2\times10^{-2}\text{ m} to 2 s.f.

Part (c): Time for a quarter revolution

The full period of the circular motion is T=2πmpB2QT=\dfrac{2\pi m_p}{B_2Q}. A quarter revolution therefore takes t=T/4=πmp2B2Qt=T/4=\dfrac{\pi m_p}{2B_2Q}:

mpB2Q=1.67×10−273.2×10−20=5.219×10−8 s\frac{m_p}{B_2Q}=\frac{1.67\times10^{-27}}{3.2\times10^{-20}}=5.219\times10^{-8}\text{ s} t=π2×5.219×10−8=1.5708×5.219×10−8=8.197×10−8 st=\frac{\pi}{2}\times5.219\times10^{-8}=1.5708\times5.219\times10^{-8}=8.197\times10^{-8}\text{ s}

Recomputing independently using t=2πr4v=πr2vt=\dfrac{2\pi r}{4v}=\dfrac{\pi r}{2v} with the radius from part (b): t=π×4.175×10−22×8.0×105=0.13121.6×106=8.20×10−8 st=\dfrac{\pi\times4.175\times10^{-2}}{2\times8.0\times10^{5}}=\dfrac{0.1312}{1.6\times10^{6}}=8.20\times10^{-8}\text{ s}, the same answer both ways (small rounding only), confirming t≈8.2×10−8 st\approx8.2\times10^{-8}\text{ s} to 2 s.f.

Final answers

  • (a) v=8.0×105 m s−1v=\boxed{8.0\times10^{5}}\text{ m s}^{-1}
  • (b) r=4.2×10−2 mr=\boxed{4.2\times10^{-2}}\text{ m}
  • (c) t=8.2×10−8 st=\boxed{8.2\times10^{-8}}\text{ s}