Questions
A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of -particle (mass 4 m and charge + 2 e) so that it can revolve in the path of same radius
detailed solution
Correct option is A
we know that r=mvqB or r2=m2v2q2B2In terms of energy, the above equation can be written asr2=2m×1/2mv2q2B2=2mEq2B2Here r and B are same, hence m1E1q12=m2E2q22 or m×1e2=3m×E24e2∴ E2=1MeVTalk to our academic expert!
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A charged beam consisting of electrons and positrons enter into a region of uniform magnetic field B perpendicular to field, with a speed v. The maximum separation between an electron and a positron will be (mass of electron = mass of positron = m)
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