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Difficulty: HardConduction of Electricity Through Gases and Cathode Rays

An electron beam traveling horizontally enters a region containing mutually perpendicular uniform electric and magnetic fields. The electric field intensity is 4.4×103 V m−14.4 \times 10^3\text{ V m}^{-1} and the magnetic flux density is 5.0×10−4 T5.0 \times 10^{-4}\text{ T}. If the beam passes through undeflected and subsequently enters a region containing only the magnetic field BB, what is the radius of the circular path executed by the electrons? (Take the specific charge of an electron em=1.76×1011 C kg−1\frac{e}{m} = 1.76 \times 10^{11}\text{ C kg}^{-1})

  1. 0.10 m0.10\text{ m}Answer
  2. B
    0.88 m0.88\text{ m}
  3. C
    10.0 m10.0\text{ m}
  4. D
    0.20 m0.20\text{ m}

Answer

The radius of the circular path executed by the electrons in the magnetic field is 0.10 m0.10\text{ m}.
Under velocity selection conditions, equal electric and magnetic forces (eE=evBeE = evB) establish electron speed v=EB=8.8×106 m s−1v = \frac{E}{B} = 8.8 \times 10^6\text{ m s}^{-1}. When moving solely through magnetic field BB, magnetic force provides centripetal force (evB=mv2revB = \frac{mv^2}{r}), yielding orbital radius r=v(e/m)B=8.8×1061.76×1011×5.0×10−4=0.10 mr = \frac{v}{(e/m)B} = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = 0.10\text{ m}.

Step-by-Step Solution

1
Calculate the electron velocity using the undeflected crossed fields condition (velocity selector)
v=EB=4.4×103 V m−15.0×10−4 T=8.8×106 m s−1v = \frac{E}{B} = \frac{4.4 \times 10^3\text{ V m}^{-1}}{5.0 \times 10^{-4}\text{ T}} = 8.8 \times 10^6\text{ m s}^{-1}
When electric and magnetic forces are equal and opposite (eE=evBeE = evB), the electrons travel in a straight line undeflected.
2
Equate magnetic force to centripetal force in the region with magnetic field only
evB=mv2r  ⟹  r=mveB=v(em)BevB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} = \frac{v}{\left(\frac{e}{m}\right)B}
The magnetic Lorentz force provides the necessary centripetal force for circular motion.
3
Substitute numerical values to find the radius rr
r=8.8×1061.76×1011×5.0×10−4=8.8×1068.8×107=0.10 mr = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = \frac{8.8 \times 10^6}{8.8 \times 10^7} = 0.10\text{ m}
Direct algebraic simplification yields the orbital radius in meters.

Key Concept

Deflection of cathode rays (electrons) in crossed electric and magnetic fields (velocity selector) and circular orbital dynamics in uniform magnetic fields.
Estimated Time:2m 0s
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Topics

Conduction of Electricity Through Gases and Cathode Rays

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