A wire loop of radius 12 cmand resistance$\mathbf{8}\mathbf{.}\mathbf{5}\mathit{\Omega }$is located in a uniform magnetic field $\overrightarrow{\mathbf{B}}$that changes in magnitude as given in Figure. The vertical axis scale is set by${\mathit{B}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathit{T}$, and the horizontal axis scale is set by${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathit{s}$. The loop’s plane is perpendicular to${\mathit{B}}_{\mathbf{s}}$. What emf is induced in the loop during time intervals (a) 0 to 2.0 s,(b) 2.0 s to 4.0 s, and (c) 4.0 s to 6.0 s?

i) The wire loop of radius is 12 cm .

ii) The resistance is,$R=8.5\Omega$

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:

$$\begin{gathered} {\Phi }_B=BA \\ \epsilon =-\frac{d{\Phi }_B}{dt} \end{gathered}$$

Where,${\Phi }_B$is magnetic flux, B is magnetic field, A is area, 𝜀 is emf.

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

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

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

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

Therefore,

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

But, substituting $A=\tau \tau r^2$,

$\epsilon =-\tau \tau r^2\frac{dB}{dt}$

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

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

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

$\epsilon =-1.1\times {10}^{2}V$

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" $$\begin{gathered} \epsilon =-\tau \tau r^2\frac{dB}{dt} \\ \epsilon =-\tau \tau {\left(0.12m\right)}^2\left(0\right) \\ \epsilon =0 \end{gathered}$$

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

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

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

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

$\epsilon =1.1\times {10}^{2}V$

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