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NEET 2001 (AIPMT) · Cathode Rays, Thomson's Experiment & Gas Discharge

Cathode Rays, Thomson's Experiment & Gas Discharge question from NEET 2001 (AIPMT)

Q7NEET 2001 · AIPMTMedium

Cathode Rays, Thomson's Experiment & Gas Discharge

In Thomson mass spectrograph E⃗⊥B⃗\vec{E} \perp \vec{B}then the velocity of electron beam will be

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Answer: (a)

  1. Undeflected beam: electric and magnetic forces balance, eE=evBeE = evB.
  2. v=∣E⃗∣∣B⃗∣v = \frac{|\vec{E}|}{|\vec{B}|}
  3. So the answer is (a) ∣E⃗∣∣B⃗∣\frac{|\vec{E}|}{|\vec{B}|}.
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