An electron with charge e enters a uniform magnetic field `vecB` with a velocity `vecv`.The velocity is perpendicular to the magnetic field. The force on the charge e is given by F= Bev. Obtain the dimensions of `vecB`.

A charge q enters into a magnetic field (B) perpendicularly with velocity V. The time rate of work done by the magnetic force on the charge is

A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

The path of a charged particle in a uniform magnetic field, when the velocity and the magnetic field are perpendicular to each other is a

An electron (charge q coulomb) enters a magnetic field of H "weber"//m^(2) with a velocity of vm//s in the same direction as that of the field the force on the electron is

If an electron is going in the direction of magnetic field vecB with the velocity of vecv then the force on electron is

A n electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity v then it travels

A proton charge (+ e coulomb) enters in a magnetic field of strength B (Tesla) perpendicular to the magnetic lines of force, with speed v. The force on the proton is -

Statement - 1 : For a charged particle to pass through a uniform electro-magnetic field without change in velocity, its velocity vector must be perpendicular to the magnetic field. Statement - 2 : Net Lorentz force on the particle is given by vec(F)=q[vec(E)+vec(v)xxvec(B)] .