A magnetic field of `(4.0xx10^-3hatk)T` exerts a force `(4.0hati+3.0hatj)xx10^-10N` on a particle having a charge `10^-9C` and moving in te `x-y` plane. Find the velocity of the particle.

A magnetic field of 8hatkmT exerts a force of (4.0hati+3.0hatj)xx10^(-10)N on a particle having a charge of 5xx10^(-10)C and going in the X-Y plane.Find the velocity of the particle.

A magnetic field of (4.0 X10^(-3)vec k) T exerts a force of (4.0veci +3.0vecj x 10^(-10) N on a particle having a charge of 1.0X10^(-9) C and going in the x-y plane. Find the velocity of the particle.

A magnetic field 4xx10^(-3) kT exerts a force (4 hat I + 3 hat j) xx10^(10) N on a particle having a charge 10^(-9) C and going on the XY plane The velocity of the particle is

A electron experiences a force (4.0hati +3.0hatj) xx 10^(-13)N in a uniform magnetic field when its velocity is 2.5hak xx 10^(7)ms^(-1) When the velocity is redirected and becomes (1.5hati - 2.0hatj)xx10^(7) xx 10^(-7) the magnetic force of the electron is zero. The magnetic field vectro oversetrarr(B) is .

A charged particle carrying charge 1 mu C is moving with velocity (2 hati + 3hati + 4hatk) ms^(-1) . If an external magnetic field of (5 hati + 3hatj - 6hatk) xx 10^(-3)T exists in the region where the particle is moving then the force on the particle is vecF xx 10^(-9)N . The vector vecF is :

A beam of protons enters a uniform magnetic field of 0.3T with velocity of 4xx10^5m//s in a direction making an angle of 60^(@) with the direction of magnetic field. The path of motion of the particle will be

A particle moves in the x-y plane such that its coordinates are given x = 10sqrt(3) t, y = 10t - t^(2) and t is time, find the initial velocity of the particle.

A charge of mass 50 mg and carrying 5xx 10^(-9)C is approaching a fixed charge 10 xx 10 ^(-9) C with a velocity point 0.5 m//s .Find the maximum approch of the charge .

An experimenter's diary reads as follows, "a charged particle is projected in a magnetic field of (7.0hati-3.0hatj)xx10^(-3)T .The acceleration of the particle is found to be (xhati+7.0hatj)xx10^(-6)m//s^(2) .Find the value of x .