Question

The following experiment was set up:

12 5 Fig.5.1 shows a rigid, straight metal rod XY placed perpendicular to a magnetic field. The magnetic field is produced by two magnets that are placed on a U-shaped steel core. The steel core sits on a digital balance steel magnets balance Fig.5.1 The weight of the steel core and the magnets is 2.500N. The rod is clamped at points X and Y The rod is connected to a battery, switch and ammeter as shown in Fig. 5.1. The direction of the magnetic field is perpendicular to the rod. Switch S is closed

Discuss what you might see if a strong field or large current is used. ( That is, jumping of the wire in what direction and why; in essence relate to Flemings left hand rule)

Answer 1

According to Lofacuting force equation, changes in motion conference magentic force when placed in magnetic field the force is given by F = q (V times B) = q (VB sin theta) V is velocity of change B is magnetic field theta is angle between V and B q is change In a current carrying conductor, changes move with drift velocity v_d Ths 1) F = q v_d B sin theta. if N be the number of changes in wire of length l, F = N (q v_d B sin theta). also N = nv that is number of electrons in wire is equal to (number of changes for unit column) (volume) or N = n times v if A is the area of correction of wire then v = Al using 2) in 1) we get F = nAl (q v^d sin theta). again n A q v_d = I (current)