1.) Magnetic information on hard drives is accessed by a read head that must mov

Question

1.) Magnetic information on hard drives is accessed by a read head that must move rapidly back and forth across the disk. The force to move the head is generally created with a voice coil actuator, a flat coil of fine wire that moves between the poles of a strong magnet, as in the figure. Assume that the coil is a square 1.0 cm on a side made of 210 turns of fine wire with total resistance 1.4 ? . The field between the poles of the magnet is 0.26 T ; assume that the field does not extend beyond the edge of the magnet. The coil and the mount that it rides on have a total mass of 14 g

Part A

If a voltage of 5.0 V is applied to the coil, what is the current? Express your answer to two significant figures and include the appropriate units.

Part B

If the current is clockwise viewed from above, what is the direction of the net force on the coil? If the current is clockwise viewed from above, what is the direction of the net force on the coil?

A. Directed up in the "Top view" figure.

B. Directed down in the "Top view" figure.

C. Directed to the left in the "Top view" figure.

D.Directed to the right in the "Top view" figure.

Part C

What is the magnitude of the net force on the coil? Express your answer to two significant figures and include the appropriate units.

Part D

What is the magnitude of the acceleration of the coil? Express your answer to two significant figures and include the appropriate units.

Part E

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval ?t from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is B.

What is the magnitude of the average emf induced in the coil?

If a symbolic answer is required, your expression must include B, A, and any other variables.

Explanation / Answer

A) V = I R

5 = I ( 1.4)

I = 3.57 A

B) Force, F = IL X B

B is into the page and IL is downward.
force will be to the right.

C) F = N ( IL x B) = 210 x 3.57 x 0.01 x 0.26 = 1.95 N

d) a = F / m = 1.95 / (14 x 10^-3 kg) = 139.3 m/s^2

e) induced emf = rate of change of flux = d(NBA)./dt

   = N A B / deltaT

(in deltaT times flux change from NAB to 0 )

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