A charged particle +q in an electric field E experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field B . But this magnetic force is perpendicular to both velocity v of the charged particle and the magnetic field B, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses m and having charges − q and +2q respectively. They are accelerated from rest through the same potential difference V and acquire kinetic energy K$_{1}$ and $K^2$. Then they enter in a region of uniform magnetic field B perpendicular to their velocities.

(i) The ratio of their kinetic energies 1 is : 1 1 2 B = B$_{0}$ k̂ v$_{1}$ = v$_{1}$ î v$_{2}$ = v$_{2}$ î

(ii) The ratio of the radii of the circular paths described by them 1

(iii) Suppose particles 1 and 2 enter the magnetic field B = B$_{0}$ k with velocities v$_{1}$ = v$_{1}$ ^ ^ i and v$_{2}$ = v$_{2}$ i . Then : 1 (A) both particles revolve clockwise (B) both particles revolve anticlockwise (C) particle 1 revolves clockwise while particle 2 revolves anticlockwise (D) particle 1 revolves anticlockwise while particle 2 revolves clockwise

(iv)

(v)

(a) If period of revolution for particle 1 is 4 s, then for particle 2, the period will be : 1

(b) If the value of momentum for particles 1 and 2 are p$_{1}$ and p$_{2}$, then : 1 (B) p$_{1}$ = p$_{2}$ (C) p$_{1}$ = 2p2 (D) p$_{1}$ = 4p2 R (u)