Question

Experiments to study vision often need to track the movements BI0 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 with a coil of very fine wire circling the edge. A current is induced in the coil each time the subject rotates his eye. Consider the experiment of FIGURE P30.60 in which a 20 -turn, 6.0-mm-diameter coil of wire circles the subject's cornea while a 1.0 T magnetic field is directed as shown. The subject begins by looking straight ahead. What $\text{emf}$is induced in the coil if the subject shifts his gaze by $5^{\circ }\text{in}0.20\mathrm{s}?$

Short answer

The $\text{emf}$induced in the coil is,

$\epsilon =0.25\mathrm{mV}$

Step-by-step solution

Magnetic Flux

The magnetic flux is the measure of magnetic force that goes through the coil of area A. The magnetic force is given by, when the magnetic force is parallel to the plane's normal.

${\mathrm{\Phi }}_m=NBA$

The magnetic flux given by,

${\mathrm{\Phi }}_{\mathrm{m}}=NBA\cos \theta$

The radius of the coil's diameter is d = 6mm, hence the coil's area equals

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{6\times {10}^{-3}\mathrm{m}}{2}\right)}^2=28.3\times {10}^{-6}{\mathrm{m}}^2$

Faraday's law

The induced, according Faraday's law, is the change in magnetic moment inside the loop.

$\epsilon =\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}$

$=\frac{dNBA\cos \theta }{dt}$

$=NBA\frac{d\cos \theta }{dt}$

$=\frac{NBA\sin \theta }{dt}$

Plug the values

we plug the values for,$\theta ,N,B,A\text{and}dt=0.20\mathrm{s}$

$\epsilon =\frac{NBA\sin \theta }{dt}$

$=\frac{\left(20\right)\left(28.3\times {10}^{-6}{\mathrm{m}}^2\right)\left(1\mathrm{T}\right)\left(\sin 5^{\circ }\right)}{\left(0.2\mathrm{s}\right)}$

$=0.25\times {10}^{-3}\mathrm{V}$

$=0.25\mathrm{mV}$