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MHT CET 2023 (9 May, Shift 1) · Other / Mixed

Other / Mixed question from MHT CET 2023 (9 May, Shift 1)

Q1MHT CET 2023 · 9 MAY · SHIFT 1Easy

Other / Mixed

An electron (mass mm) is accelerated through a potential difference of 'VV' and then it enters in a magnetic field of induction 'BB' normal to the lines. The radius of the circular path is (e=e = electronic charge)

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Answer: (b)

  1. After acceleration: 12mv2=eV⇒v=2eV/m\frac{1}{2}mv^2 = eV \Rightarrow v = \sqrt{2eV/m}.
  2. In the field: r=mveB=1B2mVer = \frac{mv}{eB} = \frac{1}{B}\sqrt{\frac{2mV}{e}}
  3. r=2VmeB2r = \sqrt{\frac{2Vm}{eB^2}}
  4. So the answer is (b).
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