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Difficulty: MediumConduction of Electricity Through Gases and Cathode Rays

A narrow beam of cathode rays is produced by accelerating electrons from rest through an unknown potential difference VV. The beam then enters a region containing mutually perpendicular uniform electric and magnetic fields of magnitudes 1.2×104 V/m1.2 \times 10^4\text{ V/m} and 3.0×10−3 T3.0 \times 10^{-3}\text{ T}, respectively. If the cathode rays pass through the crossed fields completely undeflected, what is the value of the accelerating potential difference VV? [Take electron charge e=1.6×10−19 Ce = 1.6 \times 10^{-19}\text{ C} and electron mass m=9.0×10−31 kgm = 9.0 \times 10^{-31}\text{ kg}]

  1. 45 V45\text{ V}Answer
  2. B
    90 V90\text{ V}
  3. C
    180 V180\text{ V}
  4. D
    7.2×10−18 V7.2 \times 10^{-18}\text{ V}

Answer

The accelerating potential difference VV is 45 V45\text{ V}.
For cathode rays passing undeflected through crossed fields, the electric force equals the magnetic force (eE=evBeE = evB), giving speed v=E/B=4.0×106 m/sv = E/B = 4.0 \times 10^6\text{ m/s}. The work done by the electric field during acceleration from rest equals the kinetic energy: eV=12mv2eV = \frac{1}{2}mv^2. Rearranging for VV yields V=mv22e=(9.0×10−31)(4.0×106)22×(1.6×10−19)=45 VV = \frac{m v^2}{2e} = \frac{(9.0 \times 10^{-31}) (4.0 \times 10^6)^2}{2 \times (1.6 \times 10^{-19})} = 45\text{ V}. Therefore, the option stating 45 V45\text{ V} is correct.

Step-by-Step Solution

1
Calculate the velocity vv of the cathode ray electrons using the undeflected condition in crossed fields
v=EB=1.2×104 V/m3.0×10−3 T=4.0×106 m/sv = \frac{E}{B} = \frac{1.2 \times 10^4\text{ V/m}}{3.0 \times 10^{-3}\text{ T}} = 4.0 \times 10^6\text{ m/s}
When cathode rays pass undeflected through perpendicular electric and magnetic fields, the electric force Fe=eEF_e = eE balances the magnetic force Fb=evBF_b = evB.
2
Relate the kinetic energy gained by the electrons to the accelerating potential difference VV
eV=12mv2  ⟹  V=mv22ee V = \frac{1}{2} m v^2 \implies V = \frac{m v^2}{2 e}
The work done by the accelerating potential difference equals the final kinetic energy of the electrons starting from rest.
3
Substitute the known physical quantities into the potential difference formula
V=(9.0×10−31 kg)×(4.0×106 m/s)22×(1.6×10−19 C)=9.0×10−31×1.6×10133.2×10−19=45 VV = \frac{(9.0 \times 10^{-31}\text{ kg}) \times (4.0 \times 10^6\text{ m/s})^2}{2 \times (1.6 \times 10^{-19}\text{ C})} = \frac{9.0 \times 10^{-31} \times 1.6 \times 10^{13}}{3.2 \times 10^{-19}} = 45\text{ V}
Evaluating the expression yields the required voltage V=45 VV = 45\text{ V}.

Key Concept

Deflection of Cathode Rays in Crossed Fields and Energy Conversion in Cathode Ray Tubes
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Topics

Conduction of Electricity Through Gases and Cathode Rays

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