0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

A particle of specific charge qm=π Ckg1 is projected from the origin toward positive xaxis with a velocity of 10 ms1 in a uniform magnetic field B=2^kT. The velocity v of particle after time t=112 s will be (in ms1)
  1. 5[3^i^j]
  2. 5[^i+3^j]
  3. 5[3^i+^j]
  4. 5[^i+^j]

A
5[^i+3^j]
B
5[3^i^j]
C
5[3^i+^j]
D
5[^i+^j]
Solution
Verified by Toppr

Time period T=2πmqB=2πqmB=2ππ×2=1s

Since the particle will be at point P after time t=112=T12s , it is deviated by an angle θ=2π12=300

Hence, velocity at point P

v=10(cos300^i+sin300^j)=10(32^i+12^j)=5(3^i+1^j)=5(3^i+^j)

190841_166411_ans.png

Was this answer helpful?
3
Similar Questions
Q1
A particle of specific charge qm=π Ckg1 is projected from the origin toward positive xaxis with a velocity of 10 ms1 in a uniform magnetic field B=2^kT. The velocity v of particle after time t=112 s will be (in ms1)
View Solution
Q2
A charged particle of specific charge α is released from origin at time t=0 with velocity v=v0(^i+^j) in magnetic field B=B0 ^i. The coordinate of the particle at time t=πB0α will be
View Solution
Q3
Let a=^i+^j+^k, b=^i^j+^k and c=^i^j^k be three vectors. A vector v in the plane of a and b, whose projection on c is 13, is given by
View Solution
Q4
A charged particle of specific charge, α=(qm) is released at origin at t=0 with velocity v=v0(^i+^j) in a uniform magnetic field B=B0^i. Coordinates of particle at time t=πB0α are:
View Solution
Q5
A charged particle of specific charge α is released from origin at time t=0 with velocity v=v0(^i+^j) in a uniform magnetic field B=B0^i. Find the coordinates of the particle at time t=πB0α are [α=(q/m)]
View Solution