A particle of charge +q and mass m moving under the influence of a uniform electric field E^i and uniform magnetic field B^k follows a trajectory from P to Q as shown in figure. The velocities at P and Q are v^i and −2v^j. Among the following the correct statements is/are :
Net work done = Change in Kinetic Energy
Therefore, net work done from P to Q=32mv2
And all work is done by electric field therefore equating
qE(2a)=32mv2
⇒E=3mv24qaTherefore, option A is correct.
Option B is wrong.
Using Dimensional Analysis of given option is of force not of power.
Option D is correct.
Because at Q, electric force is perpendicular to velocity vector.
⇒E=3mv24qa
Option B is wrong.
Using Dimensional Analysis of given option is of force not of power.
Option D is correct.
Because at Q, electric force is perpendicular to velocity vector.
A particle of charge +q and mass m moving under the influence of a uniform electric field E^i and a uniform magnetic field B^k follows trajectory from P to Q as shown in figure. The velocities at P and Q are v^i and −2v^j respectively. Which of the following statement(s) is/are correct
Magnetic field in the cylindrical region of radius R as shown in figure is given as B=B0t. A charge q is taken from O to D via various path, then
A charged particle of mass m and charge q, moving under the influence of a uniform electric field E ^i and a uniform magnetic field B ^k, follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, v^i and −2v^j. Then, which of the following statements (A, B, C, D) are correct? (Trajectory shown is schematic and not to scale)
(A) E=34(mv2qa)
(B) Rate of work done by the electric field at P is 34(mv2a).
(C) Rate of work done by both the fields at Q is zero.
(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.