Questions
A proton of energy 8 e V is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be
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Correct option is C
r=2mKqB ⇒q ∝ mK ⇒ K ∝ q2m⇒KαKp = qαqp2 × mpmα ⇒ Kα8 = 2qpqp2 × mp4mp=1⇒ Kα = 8 eVSimilar Questions
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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