Electric field strength `bar(E)=E_(0)hat(i)` and `bar(B)=B_(0)hat(i)` exists in a region. A charge is projected with a velocity `bar(v)=v_(0)hat(j)` at origin , then

Electric field strength bar(E)=E_(0)hat(i) and magnetic induction bar(B)=B_(0)hat(i) exists in a region. A charged particle 'q' is released from rest at origin. Work done by both the fields is after certain time is

Electric field and magnetic field in a region of space are given by (Ē= E_0 hat j) . and (barB = B_0 hat j) . A charge particle is released at origin with velocity (bar v = v_0 hat k) then path of particle is

A uniform electric field of strength e exists in a region. An electron (charge -e, mass m) enters a point A with velocity V hat(j) . It moves through the electric field & exits at point B. Then :

An electron moving with a velocity V_(1)=hat im/s at a point in a magnetic field experiences a force bar(F)= (-ehat j) Newton where 'e' is the charge of the electron.If the electron moves with a velocity bar(V)_(2)=2 hat k m/s at the same point the force experienced by it is

In a certain region static electric and magentic fields exist. The magnetic field is given by vec(B) = B_(0) (hat(i) + 2hat(j)-4hat(k)) . If a test charge moving with a velocity, vec(upsilon) = upsilon_(0)(3hat(i)-hat(j) +2hat(k)) experience no force in that region, then the electric field in the region, in SI units, is-

An electron moving with a velocity V_(1)=hat im/s at a point in a magnetic field experiences a force bar(F)=(-ehat j) Newton where 'e' is the charge of the electron.If the electron moves with a velocity bar(V)_(2)=2hat km/s at the same point the force experienced by it is

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)

Electric and magnetic field are directed as E_(0) hat(i) and B_(0) hat(k) , a particle of mass m and charge + q is released from position (0,2,0) from rest. The velocity of that particle at (x,5,0) is (5 hat(i) + 12 hat(j)) the value of x will be