Home
Class 12
PHYSICS
A particle is moving with velocity vec(v...

A particle is moving with velocity `vec(v)=hat(i)+3hatj` and it produces an electrostatic field at a point given by `vec(E)=2hat(k)`. It will produce magnetic field at that point equal to (all quantities are in S.I. units and speed of light is `c`)

A

`(6hati-2hatj)/c^(2)`

B

`(6hati+2hatj)/c^(2)`

C

zero

D

can not be determined from the given data

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • ELECTRODYNAMICS

    RESONANCE ENGLISH|Exercise Exercise-2 PART-2|26 Videos
  • ELECTRODYNAMICS

    RESONANCE ENGLISH|Exercise Exercise-2 PART-3|18 Videos
  • ELECTRODYNAMICS

    RESONANCE ENGLISH|Exercise Exercise-1 PART-2|54 Videos
  • ELECTRO MAGNETIC WAVES

    RESONANCE ENGLISH|Exercise Exercise 3|27 Videos
  • ELECTROMAGNETIC INDUCTION

    RESONANCE ENGLISH|Exercise A.l.P|19 Videos

Similar Questions

Explore conceptually related problems

A particle is moving wirh velocity vecv=hati+3hatj and it produces an electric field at a point given by vecE=2hatk . It will produce magnetic field at that point equal to (all quantities are in SI units)

A charge particle is moving with a velocity 3hat(i) + 4 hat(j) m//sec and it has electric field E = 10hat(k) N//C at a given point. Find the magnitude of magneitc field at the same point due to the motion of the charge particle.

A particle is moving with speed 6m//s along the direction of vec(A)=2hat(i)+2hat(j)-hat(k) , then its velocity is :

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

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 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 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 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)

A "bar" magnet of moment vec(M)=hat(i)+hat(j) is placed in a magnetic field induction vec(B)=3hat(i)+4hat(j)+4hat(k) . The torque acting on the magnet is

Find the potential V of an electrostatic field vec E = a(y hat i + x hat j) , where a is a constant.