Pls solve 23q iii) â ‹Q23 A proton moving towards east in a horizontal plane enters a horizontal - Physics - Moving Charges And Magnetism

Question

Pls solve 23q iii)

â€‹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)

Aarushi Mishra · Accepted Answer

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→×B→mproton=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 figure∆CPQ 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