Pls solve 23q iii)

​Q23. A proton moving towards east in a horizontal plane enters a horizontal magnetic field of 0.34 T directed towards north with speed of 2.0 × 107 ms–1. Calculate (i) the magnitude and direction of the force on the proton, (ii) radius of proton's path and (iii) the lateral displacement of the proton while moving 0.20 m towards east.
                                                                            Ans. (i) 1.09 × 10–12 N, vertically upwards, 

                                                                                    (ii) 0.625 m, (iii) 0.032 m (approx).

We know that in a uniform magnetic field,a charge moving with some velocity perpendicular to magnetic field follows a circle.Let the north direction be j^ and the east direction be i^v=2.0×107 i^ m/sB=0.34j^Tq=e=1.602×10-19CF=qv×Bmproton=1.674×10-27kgF=1.602×10-192.0×107 i^×0.34j^=1.09×10-12k^  Note: k^ points vertically upwards R=mvqB=1.674×10-27×2×1071.602×10-19×0.34=0.625m

We need to find the lateral displacement of the object when it has gone 0.2m east. We need to find the distance of particle from horizontal axis when its distance from verticlal axis k^ direction is 0.2mSee the figureCPQ is a right angle triangleBy pythagoras theoremCP2=CQ2+QP2CQ2=CP2-QP2CQ2=R2-0.22=0.6252-0.22=0.351CQ=0.351=0.592mLateral displacement=Distance of particle from horizontal axis=R-CQ=0.625-0.591=0.032m


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