91影视

i) The wire loop of radius is 12 cm .

ii) The resistance is,$R=8.5惟$

iii) The uniform magnetic field is perpendicular to loop plane.

iv) Fig30-35.

v) The vertical axis scale set by$B_s=0.50T$ .

vi) Horizontal scale is set by$t_s=6.00s$ .

Substituting Eq.30-2 in 30-4, find the equation for the emf induced due to the change in the magnetic flux. Now, applying the given intervals in Fig30-35, find. Using this value, find theemf induced in the loop during the given time intervals.

Faraday's law of electromagnetic induction states, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formula e are as follows:

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

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

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

${桅}_B=BA....................................................................\left(30-2\right)$

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

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

Therefore,

$$\begin{gathered} 蔚=-\frac{d\left(BA\right)}{dt} \\ 蔚=-A\frac{dB}{dt} \end{gathered}$$

But, substituting $A=蟿蟿r^2$,

$蔚=-蟿蟿r^2\frac{dB}{dt}$

For$0<t<2.0s$ , from Fig.30-35, there is change in B with respect to t . Thus,

$$\begin{gathered} 蔚=-蟿蟿r^2\frac{dB}{dt} \\ 蔚=-蟿蟿{\left(0.12m\right)}^2\left(\frac{0.5T-0}{2.0s-0}\right) \\ 蔚=-1.1脳{10}^{2}V \end{gathered}$$

Hence, the emf induced in the loop during time intervals for $0<t<2.0s$is ,

$蔚=-1.1脳{10}^{2}V$

For$2.0S<t<4.0s$ , from Fig.30-35, there is no change in B with respect to t . That is, B is constant. Thus,

role="math" localid="1661834120324" $$\begin{gathered} 蔚=-蟿蟿r^2\frac{dB}{dt} \\ 蔚=-蟿蟿{\left(0.12m\right)}^2\left(0\right) \\ 蔚=0 \end{gathered}$$

Hence, the emf induced in the loop during time intervals for $2.0s<t<4.0s$is, $蔚=0$

For$4.0s<t<6.0s$ , from Fig.30-35, there is change in B with respect to t . Thus,

$$\begin{gathered} 蔚=-蟿蟿r^2\frac{dB}{dt} \\ 蔚=-蟿蟿{\left(0.12m\right)}^2\left(\frac{0-0.5T}{6.0s-4.0s}\right) \\ 蔚=1.1脳{10}^{2}V \end{gathered}$$

Hence, the emf induced in the loop during time intervals for $4.0s<t<6.0s$is,

$蔚=1.1脳{10}^{2}V$

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