A charged particle of mass ‘m’ and charge ‘q’ moving under the influence of uniform electric field Ei→ and a uniform magnetic field Bk→ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi→ and −$$2vj$$→. Then which of the following statements (A$$,B,C,$$D) are the correct ? (Trajectory$$ shown is schematic and not to $$scale)A) $$E=34mv2qaB)$$ Rate of work done by the electric field at P is $$34mv3aC)$$ Rate of work done by both the fields at Q is $$zeroD)$$ The difference between the magnitude of angular momentum of the particle at P and Q is $$2$$ mav

Question

A charged particle of mass ‘m’ and charge ‘q’ moving under the influence of uniform electric field  Ei→ and a uniform magnetic field  Bk→ follows a trajectory from point P to Q as shown in  figure. The velocities at P and Q are respectively, vi→  and −$$2vj$$→. Then which of the following statements (A$$,B,C,$$D) are the correct ? (Trajectory$$ shown is schematic and not to $$scale)A) $$E=34mv2qaB)$$ Rate of work done by the electric field at P is $$34mv3aC)$$ Rate of work done by both the fields at Q is $$zeroD)$$ The difference between the magnitude of angular momentum of the particle at P and Q is $$2$$ mav

Infinity Learn · Accepted Answer

Work done by magnetic field is zero, because at all points field is perpendicular to velocity.​Work done by electric field, $$WE=$$∫$$02aqEdx$$ $$=2qEaApplying$$ work energy theorem, ​$$WB+WE=Kf$$−Ki⇒$$0$$ $$+2qEa$$ $$=12m2v2$$−$$12mv2$$​⇒$$0+qE2a=12m4$$ν$$2$$−ν$$2$$ ⇒$$E=34mv2qa$$  Rate of  work done at P is   $$Power=F$$→.v→$$=qEi^$$.$$vi^=qEv=34mv3aRate$$ of work done at Q Is  $$Power=F$$→.v→$$=qEi^$$.$$-2vj^=qEv=0$$ Angular momentum at P, L→P​​$$=mr$$→×v→$$=maj^$$×$$vi^$$​⇒L→P​​$$=mva$$−$$k^Angular$$ momentum at Q, L→Q​​$$=mr$$→×v→$$=m2ai^$$×$$2v$$−$$j^$$​​⇒L→Q​​$$=4mva$$−$$k^$$∴∣L→Q∣−∣L→P​∣$$=4mva$$−$$mva=3mva$$

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