The force on a wire is a maximum of 4.50×10 2 Nwhen placed between the pole face

Question

The force on a wire is a maximum of 4.50×102 Nwhen 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.

Part B

If the pole faces have a diameter of 17.0 cm , estimate the current in the wire if the field is 0.300 T .

Express your answer to three significant figures and include the appropriate units.

I=0.882 A

Part C

If the wire is tipped so that it makes an angle of 15.0 with the horizontal, what force will it now feel? [Hint: What length of wire will now be in the field?]

Express your answer to three significant figures and include the appropriate units.

F_rotated= ?

(ANSWER IS NOT: .0116, .0435, 4.35*10^-2, 2.93*10^-2)

Explanation / Answer

Fm = maximum force = 4.50*10-2 N

diameter = L = 17 cm

B= magnetic field = 0.300T

B) I = current = ?

maximum force in magnetic field is given by:

Fm= ILB

I = F / L*B [ sin 90 =1]

I = 4.50*10-2 / 17*10-2*0.300

I = 0.882 A

-----------------------------

c) Since the angle 15 degree with horizontal , So effective length of wire in field:

L' = L*cos 15

Force (F) = IL'*B

F= I*L*cos 15*B ( Fm= I*L*B = 4.50*10-2N)

F =  4.50*10-2*cos 15 = 0.0434N

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