Figure 30-78 shows a wire that has been bent into a circular arc of radius r = 24cm, centred at O. A straight wire OP can be rotated about O and makes sliding contact with the arc at P. Another straight wire OQ completes the conducting loop. The three wires have cross-sectional area $\mathbf{1}\mathbf{.}\mathbf{20}{\mathbf{mm}}^{\mathbf{2}}$ and resistivity $\mathbf{p}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\Omega }\mathbf{.}\mathbf{m}$, and the apparatus lies in a uniform magnetic field of magnitude B = 0.150Tdirected out of the figure. Wire OP begins from rest at angle $\mathbf{\theta }\mathbf{=}\mathbf{0}$ and has constant angular acceleration of 12rad/sec. As functions of u (in rad), find (a) the loop’s resistance and (b) the magnetic flux through the loop. (c) For what $\mathbf{\theta }$is the induced current maximum and (d)what is the maximum?

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

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’s law to find the angle $\mathit{\theta }$when current is maximum, and finally find the maximum current in the loop.

Formula:

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

By using the concept of resistivity, we can write as

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

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

So find the resistance by rearranging the equation as

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

So, by putting the value we can write as,

$R=\frac{pr\left(2+\theta \right)}{A}$

So,by putting the value we can write as,

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

So,

$R=\left(2+\theta \right)\times 3.4\times {10}^{-3}\Omega$

As we know$\varphi =\int \overrightarrow{B}.d\overrightarrow{A}$

i.e.$\varphi =BA$

Where the area of the sector of circle is

$A=\frac{1}{2}{r}^2\theta$

So flux for uniform magnetic field

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

By substituting the value we get,

$\varphi =\frac{1}{2}\times \left(0.150\right){\left(0.24\right)}^2\theta$

$\varphi =\left(4.32\times {10}^{-3}\theta \right)Wb$

According to faraday law the inducedis given by equation

$\epsilon =-\frac{d\varphi }{dt}$

But from equation (1) we can write as

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

$\epsilon =-\frac{1}{2}B{r}^2\frac{d\theta }{dt}$

But $\frac{d\theta }{dt}=\omega$

So that the emf is,

$\epsilon =-\frac{1}{2}B{r}^2\omega$

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

$\theta =\frac{1}{2}\alpha t^2$

So, $\frac{d\theta }{dt}=\omega =\alpha t$

$\alpha$is constant angular acceleration

Hence emf is

$\epsilon =-\frac{1}{2}B{r}^2\alpha t$

Now, the current flowing through the loop is,

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

By putting the value of ,

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

By differentiating with respect to t,

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

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

At this instant angle is,

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

When the current is maximum,$\omega =\alpha t$

$\omega =\alpha \sqrt{4/\alpha }$

$\omega =\sqrt{4\alpha }$

So the maximum current from equation (2) as,

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

By putting the value we can get,

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

Where$\theta =2.0rad$