Question

The force on a wire is a maximum of6.71 10^-2 N when placed between the pole faces of a magnet.The current flows horizontally to the right and the magnetic fieldis vertical. The wire is observed to "jump" toward the observerwhen 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.
A
(c) If the wire is tipped so that it makes an angle of 10.0°with the horizontal, what force will it now feel?
N

Answer 1

(a) The magnetic force,
                F_B = IdlxB

The wire jumps towards the observer, means the force on thewire is out of the plane of the paper or along the Zaxis(horizontally) taking current along the X axis andB along the Y axis.

The vector cross product dLxB is along +ve Zaxis only if the B field is vertically upward,along the Y axis. For this the upper pole must be a South pole.

(b) D= 15 cm = 0.15 m is the length of the wire in the Bfield.

The max force, F_m = BIL

          I =F_m/B.L = 6.71x10^-2/0.16x0.15

           = 2.796 A

(c) The wire is tipped at 10^o with thehorizontal means,

the angle betwen the wire(dL) andB field is 90 -10 = 80^o

The, Force, F = BIL sin 80

                      = F_m.sin 80 = 6.71x10^-2 x0.9848

                      = 6.61x10^-2N

One can also take the effective length of the wire inthefield,

                 L' = L cos 10^o

Then,          F= BIL cos 10^o

                      = F_m.cos 10^o = 6.71x10^-2x0.9848

                      = 6.61x10^-2N