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=0
the path of the particle may be x2+y2=25
the path of the particle may be y2+z2=25
the time period of the particle will be 3.14s
A
the time period of the particle will be 3.14s
B
the path of the particle may be x2+y2−4x−21=0
C
the path of the particle may be x2+y2=25
D
the path of the particle may be y2+z2=25
Open in App
Solution
Verified by Toppr
r=mvqB =√82+622 =5 Both x2+y2 and x2+y2−4x−21=0 are possible because Magnetic field is towards the positive z axis . T=2πmqB =2π2 =π Particle will move in x-y plane with radius equals to 5
Was this answer helpful?
0
Similar Questions
Q1
A charged particle of mass 2 kg and charge 2 C moves with a velocity→v=8^i+6^jm/sin a magnetic field →B=2^kT. Then
View Solution
Q2
A charged particle of mass 2kg and charge 2C moves with a velocity →v=8^i+6^jm/s in a magnetic field →B=2^kT. Then
View Solution
Q3
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:
View Solution
Q4
A charged particle of unit mass and unit charge moves with velocity →v=(8^i+6^j)m/s in a magnetic field of →B=2^kT. Choose the correct alternative(s).
View Solution
Q5
A charged particle of mass 2kg and charge 2C moves with a velocity →V=8^i+6^jm/s in a magnetic field →B=2^KT. Then