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# A charged particle of unit mass and unit charge moves with velocity of →v=(8^i+6^j) m/s in a magnetic field of →B=2^kT. Then:the path of the particle may be x2+y2−4x−21=0the path of the particle may be x2+y2=25the path of the particle may be y2+z2=25the time period of the particle will be 3.14s

A
the path of the particle may be x2+y2=25
B
the path of the particle may be y2+z2=25
C
the time period of the particle will be 3.14s
D
the path of the particle may be x2+y24x21=0
Solution
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#### r=mvqB=√82+622=5Both x2+y2 and x2+y2−4x−21=0 are possible because Magnetic field is towards the positive z axis . T=2πmqB=2π2=πParticle will move in x-y plane with radius equals to 5

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