91影视

1. The radius of the loop is, r = 12.0 cm =0.120 m
2. The plan of elastic conductor is perpendicular to the uniform magnetic field.
3. the uniform magnetic field is, B = 0.800 T
4. the loop is starts to shrink at an instantaneous rate of$\frac{\mathrm{dr}}{\mathrm{dt}}=-75.0\frac{\mathrm{cm}}{\mathrm{s}}=0.750\mathrm{m}/\mathrm{s}$

Using the equation for the magnetic flux and the given information and applying Faraday鈥檚 law, find the emf induced in the loop.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

The magnetic flux through the loop is,

$$\begin{gathered} {桅}_B=BA\cos 胃 \\ 蔚=-\frac{d{桅}_B}{dt} \end{gathered}$$

Where,${桅}_B$is magnetic flux, B is magnetic field, A is area, 饾渶 is emf.

From Eq.30-2, the magnetic flux through the loop is,

${桅}_B=BA\cos 胃...............(30-2)$

where,$\cos 胃=1$,since the magnetic field is perpendicular to theplane of the elastic conductor.

According to Faraday鈥檚 law, the emf induced due to the change in the magnetic flux is,

$蔚=-\frac{d{桅}_B}{dt}..............(30-4)$

Form Eq.30-2,

$蔚=-\frac{d\left(BA\right)}{dt}$

But the magnetic field is uniform and the area is, $A=蟿蟿r^2$. Thus,

$$\begin{gathered} 蔚=-B\frac{d\left(蟿蟿r^2\right)}{dt} \\ 蔚=-2\mathrm{蟿蟿谤叠}\frac{\mathrm{dr}}{\mathrm{dt}} \\ 蔚=-2\mathrm{蟿蟿}(0.120\mathrm{m})(0.800\mathrm{T})(-0.750\mathrm{m}/\mathrm{s}) \\ 蔚=0.452\mathrm{V} \end{gathered}$$

Hence, the magnetic flux through the loop is$蔚=0.452V$.

Therefore, by using Faraday鈥檚 law and equation, the magnetic flux through the loop can be determined.