A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field E→$$=2E0i^$$ and B→$$=B0i^$$ with a velocity v→$$=v0j^$$ . The speed of the particle will becomes $$2v0$$ after a time.

Question

A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field  E→$$=2E0i^$$ and B→$$=B0i^$$ with a velocity v→$$=v0j^$$ . The speed of the particle will becomes $$2v0$$ after a time.

Infinity Learn · Accepted Answer

E→ is parallel to B→ and V→ is perpendicular to both,  therefore path of the particle is a helix with increasing pitch .Speed of particle at any time ‘t’ is given by $$V=Vx2+Vy2+Vz2here$$,  $$Vy2+V22=V02$$   at all times because the yz components will remain same.Only the x component will increase due to electric field.After some time, it is given that   $$V=2VoUsing$$,  $$V=Vx2+Vy2+Vz2$$ ⇒$$2Vo=Vx2+Vo2$$ ⇒Vx $$=$$ $$3Vo$$ Equation of kinematics,  $$Vx=Ux+axt$$ ⇒$$3V0=q2EoMt$$⇒$$t=3MV02qEo$$ .

A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field $\vec{E} = 2 E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{i}$ with a velocity $\vec{v} = v_{0} \hat{j}$ . The speed of the particle will becomes $2 v_{0}$ after a time.

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A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field E→=2E0i^ and B→=B0i^ with a velocity v→=v0j^ . The speed of the particle will becomes 2v0 after a time.

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A particle of charge ‘q’ and mass ‘m’ starts moving from the origin under the action of an electric field $\vec{E} = 2 E_{0} \hat{i}$ and $\vec{B} = B_{0} \hat{i}$ with a velocity $\vec{v} = v_{0} \hat{j}$ . The speed of the particle will becomes $2 v_{0}$ after a time.