A particle of charge q and mass m starts moving from the origin under the action of an electric field $$E=Eoi^$$ and $$B=Boi^$$ with velocity V→$$=Voj^$$. The speed of the particle will become $$2V0$$ after a time

Question

A particle of charge q and mass m starts moving from the origin under the action of an electric field $$E=Eoi^$$  and  $$B=Boi^$$ with velocity  V→$$=Voj^$$. The speed of the particle will become $$2V0$$ after a time

Infinity Learn · Accepted Answer

Given, $$E=Eoi^$$ , $$B=Boi^$$ and $$V=Voj^$$ ∴ V→ ⊥ E→  and V→ ⊥ B→ Hence, the path is helix with speed remaining constant for the circular path in the YZ plane and uniform accelerated motion $$ax=Eqm$$ along x axis. ∴$$Vy2+Vz2=Vo2$$ .............(1) Now at any given time speed is given by,  $$V=Vx2+Vy2+Vz2$$ ⇒ $$V2=Vx2+Vo2$$     [using (1)]Given, $$V=2V0$$; and here $$Vx=axt$$ $$4Vo2=axt2+Vo2$$ ⇒$$axt2=3Vo2$$ ⇒$$t=3Voax=3VomEq$$              ∵ $$ax=Eqm$$

A particle of charge q and mass m starts moving from the origin under the action of an electric field $E = E_{o} \hat{i} a n d B = B_{o} \hat{i}$ with velocity $\vec{V} = V_{o} \hat{j}$ . The speed of the particle will become $2 V_{0}$ after a time

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle of charge q and mass m starts moving from the origin under the action of an electric field E=Eoi^ and B=Boi^ with velocity V→=Voj^. The speed of the particle will become 2V0 after a time

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

A particle of charge q and mass m starts moving from the origin under the action of an electric field $E = E_{o} \hat{i} a n d B = B_{o} \hat{i}$ with velocity $\vec{V} = V_{o} \hat{j}$ . The speed of the particle will become $2 V_{0}$ after a time