Question

Two concentric circular coils, $C_1$ and $C_2$, are placed in the XY plane. $C_1$ has $500turns$, and a radius of $1cm$. $C_2$ has $200turns$ and a radius of $20cm$. $C_2$ carries a time-dependent current $I\left(t\right)=\left(5t^2-2t+3\right)A$ where $t$ is in $s$. The emf induced in $C_1$ (in mV), at the instant $t=1s$ is $\frac{4}{x}$. The value of $x$ is ________.

Answer 1

Step 1: Given-

The number of turns in $C_1$: $n_1=500turns$

Radius of $C_1$: $r_1=1cm=1\times {10}^{-2}m$

The number of turns in $C_2$: $n_2=200turns$

Radius of $C_2$: $R_2=20cm=20\times {10}^{-2}m$

Step 2: Formula Used

Current in $C_2$: $I\left(t\right)=\left(5t^2-2t+3\right)A$

Emf induced in $C_1$ at $t=1s$: $\epsilon =\frac{4}{x}mV=\frac{4}{x}\times {10}^{-3}V$

Magnetic flux -$\varphi$

Magnetic field-$B$

The cross-sectional area of coil-$A$

Number of turns -$n$

$\begin{array}{rcl}\varphi & =& nBA\end{array}$

Current flowing in coil-$I$

Magnetic constant/permeability constant -${\mu }_o$

$B=\begin{array}{rcl}& & \frac{n{\mu }_{0}I}{2R}\end{array}$

time -$t$

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

Step 3: Calculate the value of $\frac{dI}{dt}$ at $t=1s$.

$\begin{array}{rcl}{\overline{)\frac{dI}{dt}}}_{t=1}& =& \frac{d}{dt}\left(5t^2-2t+3\right)\\ & =& 5\left(2t\right)-2\left(1\right)+3\left(0\right)\\ & =& 10t-2\\ & =& 10\times 1-2\\ & =& 8A/s\end{array}$

Step 4: Calculate the flux through first coil due to second

Use the formulas for flux and magnetic field to calculate the flux through first coil

$\begin{array}{rcl}\varphi & =& n_1{B}_2{A}_1\\ & =& \left(500\right)\left(\frac{n_2{\mu }_{0}I}{2R_2}\right)\left(\pi {r_1}^2\right)\\ & =& \left(500\right)\left(\frac{200\times 4\pi \times {10}^{-7}\times I}{2\times 20\times {10}^{-2}}\right)\left(\pi {\left(1\times {10}^{-2}\right)}^2\right)\\ & =& 9.869\times {10}^{-5}I\end{array}$

Step 5: Calculate the value of $x$

Find the emf using the formula and substitute the value of rate of current obtained at $t=1$ an the value of emf given.

$$\begin{gathered} \epsilon =\frac{d\varphi }{dt} \\ \Rightarrow \frac{4}{x}\times {10}^{-3}=\frac{d}{dt}\left(9.869\times {10}^{-5}I\right) \\ \Rightarrow \frac{4}{x}\times {10}^{-3}=9.869\times {10}^{-5}\frac{dI}{dt} \\ \Rightarrow \frac{4}{x}\times {10}^{-3}=9.869\times {10}^{-5}\times 8 \\ \Rightarrow x=\frac{4\times {10}^{-3}}{9.869\times {10}^{-5}\times 8} \\ \Rightarrow x=5 \end{gathered}$$

Hence, $x=5$.