Experiments to study vision often need to track the movements of a subject\'s ey

Question

Experiments 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 1.0 T magnetic field is directed as shown in the figure. The subject begins by looking straight ahead.(Figure 1) What emf is induced in the coil if the subject shifts his gaze by 6.0 in 0.40 s ?

The anwer is not 1.41*10^-3

Explanation / Answer

whent he person looks straight ahead, the field is perpendicular to the plane of the loop. Let q(t) be the angle the gaze shifts over as a function of time t. The magnetic flux through the loop due to the field B, as defined by normal vector, n, to the plane of the loop is:

F = B*|A|n = |B||A| cos(q(t)) where |A| is the loop's area, n is the normal vector to the plane of the loop, B is the magnetic field vector and "*" is teh dot product. The | | is the magnitude of teh vector.

Let's drop the magnitude symbols from here out and understand we are talking about magnitudes and not vectors.

Now teh induced emf E is

E = -dF/dt = -B d(A cos(q(t))/dt = BA sin(q) dq/dt

Now to compute dq/dt, convert 8 degrees to radians ---> 6 deg = 0.10472 radians

The dq/dt = 0.10472/(0.4 s) = 0.2681 rad/s

Now A = N*pi*(d/2)^2 where N = number of tunrs and d = 6x10^-3 m

A = 5.65x10^-4 m^2

Finally setting q = 8 deg

F = 1.0T*5.65x10^-4 m^2*0.2681 rad/s*sin(6 deg) = 1.5833*10^-5 volts

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