Q#3. Part A The force on a wire is a maximum of 7.80x10-2 N when placed between
ID: 1774439 • Letter: Q
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Q#3. Part A The force on a wire is a maximum of 7.80x10-2 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. What type of magnetic pole is the top pole face? north pole south pole Submit My Answers Give Up Part B If the pole faces have a diameter of 12.5 cim, estimate the current in the wire if the field is 0.282 T 1= A Submit My Answers Give Up Part C If the wire is tipped so that it makes an angle of 15.0 with the horizontal, what force wil t now feel? F= N Submit My Answers Give UpExplanation / Answer
(a) 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= 12.5 cm = 0.125 m is the length of the wire in the B field.
The max force, Fm = BIL
I =Fm/B.L = 7.80x10-2/0.282x0.125
= 2.212 A
(c) The wire is tipped at 15o with thehorizontal means,
the angle betwen the wire(dL) andB field is 90 -15 = 75o
The, Force, F = BIL sin 75
= Fm.sin 75 = 7.80x10-2 x0.9848
= 7.68x10-2N
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