The force on a wire is a maximum of 4.50×10−2 N when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on.
(a) What type of magnetic pole is the top pole face?
Northpole
Southpole
(b) If the pole faces have a diameter of 15.0 cm, estimate the current in the wire if thefield is 0.16 T.
(c) If the wire is tipped so that it makes an angle of 10.0°with the horizontal, what force will it now feel?

a) Southpole

b) 1.875 A

c) 4.43 x 10-2 N

Explanation:

a) The magnetic force,

               FB = IdlxB

As the wire jumps towards the observer,  this means that the force on the wire is out of the plane of the paper or along the Z axis(horizontally) taking current along the X axis and B along the Y axis.

The vector cross product dLxB is along +ve Z axis only if the B field is vertically upward,along the Y axis. Therefore, the top pole face must be a South pole.

b) D= 15 cm = 0.15 m is the length of the wire in the magnetic field B

The max force, Fm = BIL

         I =Fm/B.L = 4.5x10^-2/0.16x0.15      

          = 1.875 A

The current in the wire is therefore 1.875 Amps

c) The wire is tipped at 10.0° with the horizontal means,

the angle between the wire(dL) and magnetic field B is 90° -10° = 80°

The, Force, F = BIL sin 80°

                     = Fm.sin 80° = 4.5x10^-2 x0.9848

                     = 4.43 x 10-2 N

The effective length of the wire in the field,

                L' = L cos 10°

Then,          F= BIL cos 10°

                     = Fm.cos 10° = 4.5x10^-2 x0.9848

                     = 4.43 x 10-2 N

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