A proton accelerated by a potential difference `500 KV` moves though a transverse field of `0.51 T` as shown in figure. The angle `theta` through which the proton deviates from the intial direction of its motion is

A proton accelerated by a potential difference 500 kV flies through a uniform transverse magnetic field 0.1 T. The field is spread on a region of 1.0 cm thickness. The angle through which the proton gets deviated from its original direction is , (mass of proton = 1.6 xx 10^(-27) kg and charge of proton =1.6 xx 10^(19) C )

A proton is accelerated through a potential difference of 1 V. Its energy is

A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If alpha be the angle of deviation of proton from initial direction of motion , the value of sin alpha will be

A proton accelarted by a potential differnce V = 500 kV fieles through a unifrom transverse magnetic filed the induction B = 0.54 T . The field occupies a region of space d = 10 cm in thickness (Fig). Find the angle alpha through which the proton deviates from the initial direction of its motion.

A proton of mass m and charge q is accelerated by a potential difference V in a perpendicular magnetic field B occupying space t. The value of sin theta where theta is deviation of proton from initial direction is

A proton is accelerated from rest through a potential of 500 volts. Its final kinetic energy is

When a proton is accelerated from rest through a potential difference of 1000 V, its kinetic energy becomes