Home
Class 11
PHYSICS
A particle having change q and m is proj...

A particle having change `q` and `m` is projected with velocity `vec v = 2 hat i- 3 hat j` in uniform electric field `vec E = E_0. hat j` change in momentum `|Delta vec p|` during any time interval `t` is given by :

A

`sqrt(13) M`

B

`q E_0 t`

C

`(q E_0 t)/(m)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
The product of force and time interval is called impulse. Also from Newton's second law, we have
`vec F = m vec a = m (Delta vec v)/(Delta t)`
`vec F . Delta t = m Delta vec v = Delta vec p`
The force on chaege `q` due to electric field is `qE_0`
`:.` Impulse `= qE_0 t`.
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST

    A2Z|Exercise General Kinematics|30 Videos

  • KINETIC THEORY OF GASES AND THERMODYNAMICS

    A2Z|Exercise Chapter Test|29 Videos

  • MOTION IN TWO DIMENSION

    A2Z|Exercise Chapter Test|29 Videos

Similar Questions

Explore conceptually related problems

An alpha particle moving with the velocity vec v = u hat i + hat j ,in a uniform magnetic field vec B = B hat k . Magnetic force on alpha particle is

Charge Q is given a displacement vec r = a hat i + b hat j in an electric field vec E = E_1 hat i + E_2 hat j . The work done is.

A charged particle of specific charge (charge/mass) alpha released from origin at time t=0 with velocity vec v = v_0 (hat i + hat j) in uniform magnetic field vec B = B_0 hat i. Coordinates of the particle at time t= pi//(B_0 alpha) are

A charged particle moves with velocity vec v = a hat i + d hat j in a magnetic field vec B = A hat i + D hat j. The force acting on the particle has magnitude F. Then,

A portion is fired from origin with velocity vec(v) = v_(0) hat(j)+ v_(0) hat(k) in a uniform magnetic field vec(B) = B_(0) hat(j) . In the subsequent motion of the proton

A particle of charge q and mass m released from origin with velocity vec(v) = v_(0) hat(i) into a region of uniform electric and magnetic fields parallel to y-axis. i.e., vec(E) = E_(0) hat(j) and vec(B) = B_(0) hat(j) . Find out the position of the particle as a functions of time Strategy : Here vec(E) || vec(B) The electric field accelerates the particle in y-direction i.e., component of velocity goes on increasing with acceleration a_(y) = (F_(y))/(m) = (F_(e))/(m) = (qE_(0))/(m) The magnetic field rotates the particle in a circle in x-z plane (perpendicular to magnetic field) The resultant path of the particle is a helix with increasing pitch. Velocity of the particle at time t would be vec(v) (t) = v_(x) hat(i) + v_(y) hat(j) + v_(z) hat(k)

The potential field of an electric field vec(E)=(y hat(i)+x hat(j)) is

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.

The position of a particle is given by vec r = hat i + 2hat j - hat k and its momentum is vec p = 3 hat i + 4 hat j - 2 hat k . The angular momentum is perpendicular to

An electric dipole moment vec P = ( 2. 0 hat I + 3 . 0 hat j) mu C . M is placed in a usifrom electric field vec E = (3 hat I + 2 . 0hat k ) xx 10^5 N C^(-1) .