Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Magnetic information on hard drives is accessed by a read head that must move ra

ID: 1603503 • Letter: M

Question

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 250 turns of fine wire with total resistance 1.2 . The field between the poles of the magnet is 0.30 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 13 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?

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.

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. S Side view Top view

Explanation / Answer

Given that the magnitude of magnetic field is B = 0.30 T

Number of turns in the coil is N = 250

Resistance of wire is R = 1.2

Length of the square is L = 1.0 cm = 0.01 m

Total mass of coil is m = 13 g = 0.013 kg

a) The current in the coil is I = 5.0V / R = 5 / 1.2

                                       = 4.16 A     

b) The answer is option 'D'

  Directed to the right in the "Top view" figure

c)Magnetic force on the coil is F = NBIL sin

                                               = 250*0.30T*4.16A*0.01m*sin90o

= 3.12 N

d) The acceleration of the coil is a = F /m

                                               = 3.12 N / 0.013 kg

                                               = 240 m/s2

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote