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

A charged particle moves with a constant velocity (^i+^j)m/s in a magnetic field B=(2^i+3^k) and uniform electric field E=(a^i+b^j+c^k)N/C , then (assuming all quantities in S.I. unit):
  1. a=3
  2. b=3
  3. c=2
  4. a2+b2+c2=22

A
a=3
B
a2+b2+c2=22
C
b=3
D
c=2
Solution
Verified by Toppr

Since the charge particle is moving with constant velocity, its acceleration will be 0.
Fnet=0
Fel+Fmag=0
In equilibrium, electric force = - magnetic force
So, qE=q(v×B) or a^i+b^j+c^k=(3^i3^j2^k)
So, a=3,b=3,c=2and a2+b2+c2=9+9+4=22

Was this answer helpful?
0
Similar Questions
Q1
A charged particle moves with a constant velocity (^i+^j)m/s in a magnetic field B=(2^i+3^k) and uniform electric field E=(a^i+b^j+c^k)N/C , then (assuming all quantities in S.I. unit):
View Solution
Q2
A particle of mass m and charge q is thrown from origin at t=0 with velocity 2^i+3^j+4^k units in a region with uniform magnetic field B=2^i units. After time t=πmqB, an electric field E is switched on, such that particle moves on a straight line with constant speed. E may be
View Solution
Q3
In a given region a charge particle is moving under the effect of electric and magnetic field with uniform velocity v=(^i+^j^k) m/s and magnetic field is given as B=(2^i+^j2k)T. The electric field is given as?
View Solution
Q4
An electron is moving with velocity (2^i+2^j)m/s in an electric field of intensity E=^i+2^j8^kvolt/m and a magnetic field of B=(2^j+3^k) tesla. Find the magnitude of force on the electron.
View Solution
Q5
A point charge +q0 is projected in a magnetic field B=(^i+2^j3^k). If acceleration of the particle is a=(2^i+b^j+^k) then value of b will be?
View Solution