91影视

$$\begin{gathered} r=24cm=0.24m \\ A=1.20m{m}^2=1.20脳{10}^{-6}{m}^2 \\ p=1.70脳{10}^{-8}惟.m \\ B=0.150T \\ a=12rad/sec \end{gathered}$$

To solve this problem we use the concept of resistivity to find the resistance in the loop. After that we find the magnetic flux using the relation between magnetic field and flux. After that by using Faraday鈥檚 law to find the angle $\mathit{胃}$when current is maximum, and finally find the maximum current in the loop.

Formula:

$$\begin{gathered} 蠒=BA \\ 蔚=-\frac{d蠒}{dt} \\ p=\frac{RA}{I} \end{gathered}$$

By using the concept of resistivity, we can write as

$p=\frac{RA}{I}$

Where the length of the complete loop is$I=2r+胃r$

So find the resistance by rearranging the equation as

role="math" localid="1661939773879" $$\begin{gathered} R=\frac{p\left(2r+胃r\right)}{A} \\ R=\frac{pr\left(2r+胃\right)}{A} \end{gathered}$$

So, by putting the value we can write as,

$R=\frac{pr\left(2+胃\right)}{A}$

So,by putting the value we can write as,

$R=\frac{1.70脳{10}^{-8}脳0.24脳\left(2+胃\right)}{1.20脳{10}^{-6}}$

So,

$R=\left(2+胃\right)脳3.4脳{10}^{-3}惟$

As we know$蠒=鈭�\stackrel{鈫�}{B}.d\stackrel{鈫�}{A}$

i.e.$蠒=BA$

Where the area of the sector of circle is

$A=\frac{1}{2}{r}^2胃$

So flux for uniform magnetic field

$蠒=\frac{1}{2}B{r}^2胃.................................................................\left(1\right)$

By substituting the value we get,

$蠒=\frac{1}{2}脳\left(0.150\right){\left(0.24\right)}^2胃$

$蠒=\left(4.32脳{10}^{-3}胃\right)Wb$

According to faraday law the inducedis given by equation

$蔚=-\frac{d蠒}{dt}$

But from equation (1) we can write as

$蔚=-\frac{d\left({\displaystyle \frac{1}{2}}B{r}^2胃\right)}{dt}$

$蔚=-\frac{1}{2}B{r}^2\frac{d胃}{dt}$

But $\frac{d胃}{dt}=蝇$

So that the emf is,

$蔚=-\frac{1}{2}B{r}^2蝇$

Where $胃$is angular displacement and it can be represented as,

$胃=\frac{1}{2}伪t^2$

So, $\frac{d胃}{dt}=蝇=伪t$

$伪$is constant angular acceleration

Hence emf is

$蔚=-\frac{1}{2}B{r}^2伪t$

Now, the current flowing through the loop is,

$i=\frac{B{r}^2伪t}{2R}.........................................................\left(2\right)$

By putting the value of ,

$$\begin{gathered} i=\frac{1}{2\left(2+胃\right)\left(3.4脳{10}^{-3}\right)}B{r}^2伪t \\ i=\frac{1}{2\left(2+{\displaystyle \frac{1}{2}}伪t^2\right)\left(3.4脳{10}^{-3}\right)}B{r}^2伪t \\ i=\frac{1}{2\left(4+伪t^2\right)\left(3.4脳{10}^{-3}\right)}B{r}^2伪t \end{gathered}$$

By differentiating with respect to t,

$$\begin{gathered} \frac{di}{dt}=\frac{d}{dt}\left(\frac{1}{\left(4+伪t^2\right)\left(3.4脳{10}^{-3}\right)}B{r}^2伪t\right) \\ \frac{di}{dt}=\frac{B{r}^2伪\left(4-伪t^2\right)\left(3.4脳{10}^{-3}\right)}{{\left(4+伪t^2\right)}^2\left(3.4脳{10}^{-3}\right)} \\ \frac{di}{dt}=\frac{B{r}^2伪\left(4-伪t^2\right)}{{\left(4+伪t^2\right)}^2\left(3.4脳{10}^{-3}\right)} \end{gathered}$$

The induced current is at maximum when $4-伪t^2=0$or $t=\sqrt{4/伪}$

At this instant angle is,

$$\begin{gathered} 胃=\frac{1}{2}伪t^2 \\ 胃=\frac{1}{2}伪\left(\frac{4}{伪}\right) \\ 胃=2.0rad \end{gathered}$$

When the current is maximum,$蝇=伪t$

$蝇=伪\sqrt{4/伪}$

$蝇=\sqrt{4伪}$

So the maximum current from equation (2) as,

$$\begin{gathered} i=\frac{B{r}^2蝇}{2R} \\ {i}_{max}=\frac{B{r}^2\sqrt{4伪}}{2R} \end{gathered}$$

By putting the value we can get,

$i_{max}=\frac{\left(0.150\right){\left(0.24\right)}^2\sqrt{4脳12}}{2\left(2+2.0\right)\left(3.4脳{10}^{-3}\right)}$

Where$胃=2.0rad$