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

We begin with the magnetic field produced by the larger coil $$C_2$$ at its own centre (and therefore also at the centre of the smaller concentric coil $$C_1$$). For a circular coil of $$N$$ turns carrying current $$I$$, the field on the axis at the centre is given by the well-known formula

$$B=\dfrac{\mu_0 N I}{2R}$$

where $$\mu_0=4\pi\times10^{-7}\,\text{H m}^{-1}$$ is the permeability of free space and $$R$$ is the radius of the coil.

For coil $$C_2$$ we have $$N_2=200$$ turns and $$R_2=20\ \text{cm}=0.20\ \text{m}$$. Hence

$$B(t)=\dfrac{\mu_0 N_2}{2R_2}\,I(t) =\dfrac{\mu_0\,(200)}{2(0.20)}\,I(t) =\dfrac{\mu_0\,(200)}{0.40}\,I(t).$$

The smaller coil $$C_1$$ of radius $$r_1=1\ \text{cm}=0.01\ \text{m}$$ encloses an area

$$A_1=\pi r_1^2=\pi(0.01)^2=\pi\times10^{-4}\ \text{m}^2.$$

The magnetic flux through one turn of $$C_1$$ is therefore

$$\Phi(t)=B(t)\,A_1 =\left(\dfrac{\mu_0 N_2}{2R_2}\,I(t)\right)A_1.$$

Because coil $$C_1$$ has $$N_1=500$$ turns, the total flux linkage with it is

$$\Lambda(t)=N_1\Phi(t)=N_1A_1\dfrac{\mu_0 N_2}{2R_2}\,I(t).$$

Faraday’s law of electromagnetic induction states that the induced emf is the negative time derivative of this flux linkage:

$$\mathcal{E}(t)=-\dfrac{d\Lambda}{dt} =-N_1A_1\dfrac{\mu_0 N_2}{2R_2}\,\dfrac{dI}{dt}.$$

Only the current $$I(t)$$ is time dependent, so we differentiate it. The given current is

$$I(t)=5t^2-2t+3\ \text{A},$$

hence

$$\dfrac{dI}{dt}=10t-2.$$

At the required instant $$t=1\ \text{s}$$,

$$\left.\dfrac{dI}{dt}\right|_{t=1}=10(1)-2=8\ \text{A s}^{-1}.$$

Substituting all numerical values (the sign is irrelevant because we need only the magnitude of the emf), we obtain

$$|\mathcal{E}|=N_1A_1\dfrac{\mu_0 N_2}{2R_2}\,\left|\dfrac{dI}{dt}\right| =500\;\bigl(\pi\times10^{-4}\bigr)\; \dfrac{4\pi\times10^{-7}\,(200)}{2(0.20)}\; 8.$$

Simplifying step by step,

$$\dfrac{4\pi\times10^{-7}\,(200)}{2(0.20)} =\dfrac{4\pi\times10^{-7}\times200}{0.40} =\dfrac{800\pi\times10^{-7}}{0.40} =2000\pi\times10^{-7} =2\pi\times10^{-4}.$$

So

$$|\mathcal{E}|=500\;(\pi\times10^{-4})\;(2\pi\times10^{-4})\;8 =500\times2\times8\times\pi^2\times10^{-8} =8000\pi^2\times10^{-8}\ \text{volts}.$$

Combining the powers of ten,

$$8000=8\times10^{3} \quad\Rightarrow\quad 8\times10^{3}\times10^{-8}=8\times10^{-5},$$

so

$$|\mathcal{E}|=8\pi^2\times10^{-5}\ \text{V}.$$

Since $$1\ \text{mV}=10^{-3}\ \text{V}$$, we convert the emf to millivolts:

$$|\mathcal{E}|=8\pi^2\times10^{-5}\ \text{V} =8\pi^2\times10^{-5}\times10^{3}\ \text{mV} =8\pi^2\times10^{-2}\ \text{mV} =0.08\pi^2\ \text{mV}.$$

The problem statement tells us that this magnitude equals $$\dfrac{4}{x}\ \text{mV}$$, so we write

$$\dfrac{4}{x}=0.08\pi^2.$$

Isolating $$x$$ gives

$$x=\dfrac{4}{0.08\pi^2} =\dfrac{4}{\frac{8}{100}}\;\dfrac{1}{\pi^2} =50\;\dfrac{1}{\pi^2} =\dfrac{50}{\pi^2}.$$

Taking $$\pi^2\approx9.87$$, we have

$$x\approx\dfrac{50}{9.87}\approx5.1\;.$$

To the nearest integer, required by JEE’s answer key, this is simply $$5$$.

So, the answer is $$5$$.