Questions
A proton of energy E moving in a normal magnetic field in a circular path of radius R. The energy of an particle moving in the same magnetic field with same radius is
detailed solution
Correct option is A
ω=QBm, E=12m(ωR)2=Q2B2R22m∴ EαEP=(QαQP)2.(mPmα)=(2)r.(14)=1⇒ Eα=EP=ETalk to our academic expert!
Similar 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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