Question

Experiments to study vision often need to track the movements of a subject's eye. One way...

uploaded imageExperiments to study vision often need to track the movements of a subject's eye. One way of doing so is to have the subject sit in a magnetic field while wearing special contact lenses that have a coil of very fine wire circling the edge. A current is induced in the coil each time the subject rotates his eye. Consider an experiment in which 20 turn, 6.0 a-mm-diameter coil of wire circles the subject's cornea while a 1.4 T magnetic field is directed as shown in the figure. The subject begins by looking straight ahead.

What emf is induced in the coil if the subject shifts his gaze by 5.0 degrees in 0.30 s. ?
Express your answer using two significant figures.
εcoil = _________V

0 0
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Answer #1
Concepts and reason

The concept used to solve this problem is induced electromotive force (emf).

Initially, use the relation between magnetic field, the area of cornea, and the shifting angle to calculate the change in flux.

Finally, use the flux and the number of turns to calculate the induced emf in the coil.

Fundamentals

The expression for Faraday’s law of induction is given as follows:

E=N

Here, the electromotive force is , the number of turns is , the magnetic flux is , and time is .

The expression for magnetic flux is given as follows:

= BAcos e

Here, the magnetic field is , the area of cross section is , the magnetic flux is , and the angle between the area and the magnetic field is .

The expression for the area of the circular coil is as follows:

A
=
T

Here, the area of the circular coil is .

Substitute 6.0 mm
for .

=r{(3.0mm) (10m)
= T (3.0x10 m)
= 2.83x10 m²

The expression for magnetic flux is given as follows:

= BAcos e

The expression for the initial magnetic flux is as follows:

A = BA cose,

Here, the initial magnetic flux is and the initial angle of gaze is .

Substitute for , 2.83x10m
for , and for .

0 = (1.4T)(2.83x10 m²)(cos 90°)
= (1.4T)(2.83x10 m²)(0)
= 0T-m?

The expression for the final magnetic flux is as follows:

02= BA cos ,

Here, the final magnetic flux is and the final angle of gaze is .

Substitute for , 2.83x10m
for , and for .

=(1.4T)(2.83x106m²)(cos 85°)
= 3.45x10 T.m?

The expression for the change in magnetic flux is as follows:

ΔΦ = φ, -,

Substitute 3.45x10 T.mp
for and OT-m
for .

A$ = 3.45x10 T.m-OT.m?
= 3.45x10 T.m?

The expression for Faraday’s law of induction is given as follows:

E=N

Substitute for , for , and for .

(20)(3.45x10T-m)
(0.30s)
= 2.3x10-V

Ans:

The induced emf in the coil is 2.3 x 10 V
.

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