a proton of charge e and mass m enters a uniform magnetic fieldB=bî with an initial velocity v=vxî+vyj(cap).find an expression in unit-vector notation for its velocity at time t.

Dear Padyot, We know the formula for the Force experienced by a charged particle moving with some velocity in a magnetic field As F = q{ V cross B ) Where V and B are in vector notations. after Calculations we get F = {q * [(-Vy)*b]} k Where k is a unit vector in z direction. And the acceleration in z direction is given by a= (Force in z direction)/mass of the charge = (q/m) * [(-Vy)*b] We know the equation of motion as v = u + a*t And the initial velocity in z direction is zero and thus velocity after time t seconds in z direction is given by Vz = a*t Thus the velocity vector after time t seconds is given by V = Vx i + Vy j + Vz k = Vx i + Vy j - { (q/m) * (Vy)*b } k Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best STUDENT NAME !!! Regards, Askiitians Experts Adapa Bharath