Questions
The radius of the circular path or helical path followed by the test charge qo moving in magnetic field B with some velocity v is
detailed solution
Correct option is A
The circular motion of charged particle will be due to component of velocity v perpendicular to magnetic field, i.e., v sinθ∴ m(vsinθ)2r=q0(vsinθ)Bor r=m(vsinθ)q0BTalk 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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