cos B dl B.dl B.dl B B l -------(2) a a a Where, = length of the side of the rectangular loop l ab abcd. Now calculation for total current threading the loop abcd Let number of turns per unit length of the solenoid = n Then number of turns in length of the solenoid = l nl Total current threading the loop = abcd nlI----------(3) Putting the value of (2) and(3) in (1) We get Bl = or B nI Q 4. Discuss the motion of a charged particle in a uniform magnetic field with initial velocity parallel to the field, (ii) perpendicular to the magnetic field and (iii) at an arbitrary angle with the field direction. Ans. When a charged particle having charge and velocity enters a magnetic field , it q v B experiences a force F q(v The direction of this force is perpendicular to both and . The magnitude of this force v B is F = qv B sinθ Following three cases are possible: 1. When the initial velocity is parallel to the magnetic field: Here =0 , So sin0 = 0. F = qvB Thus the parallel magnetic field does not exert any force on the moving charged particle. The charged particle will continue to move along the line of force.
B.dl B.dl B dl cos B B l -------(2) a a a Where, l = length of the side ab of the rectangular loop abcd. Now calculation for total current threading the loop abcd Let number of turns per unit length of the solenoid = n Then number of turns in length l of the solenoid = nl Total current threading the loop abcd = nlI----------(3) Putting the value of (2) and(3) in (1) We get Bl = or B Q 4. Discuss the motion of a charged particle in a uniform magnetic field with initial velocity parallel to the field, (ii) perpendicular to the magnetic field and (iii) at an arbitrary angle with the field direction. Ans. When a charged particle having charge q and velocity v enters a magnetic field B , it experiences a force F q(v The direction of this force is perpendicular to both v and B . The magnitude of this force is F = qv B sinθ Following three cases are possible: 1. When the initial velocity is parallel to the magnetic field: Here =0 , So F = qvB sin0 = 0. Thus the parallel magnetic field does not exert any force on the moving charged particle. The charged particle will continue to move along the line of force. nI
physics-xii-15-16.docx