A coil having $28$ turns with an average diameter of $0.02\,m$, is placed perpendicular to the magnetic field of $8000\,T$. If the magnetic field changes to $3000\,T$ in $4\sec $. What is the magnitude of induced e.m.f in the coil?

Hint:In order to solve this question, firstly we will find the magnitude of net change in magnetic field in given time and then by using the faraday’s law of electromagnetic induction, we will calculate the magnitude of induced e.m.f.

Formula Used:
$\varepsilon = - N\dfrac{{\Delta \phi }}{{\Delta t}}$
Here, $\varepsilon $ denotes for emf, $\phi $ is the magnetic flux.
Magnetic flux is given by: $\phi = B \times A$
$A$ Is the area of the surface crossing the magnetic field lines.

Complete step by step answer:
 Let us understand the Faraday’s law of electromagnetic induction first, it states that “the magnitude of the induced emf is always equals to the rate of change of magnetic flux”
We have given, ${B_1} = 8000T$
${B_2} = 3000T \\
\Rightarrow \Delta t = 4\sec \\
\Rightarrow A = \pi \dfrac{{{d^2}}}{4} \\ $
Now, calculating
$\Delta B = 5000T \\
\Rightarrow A = \pi \times {10^{ - 4}}{m^2} \\
\Rightarrow N = 28turns \\ $
Now, we will find change in magnetic flux which is given as:
$\Delta \phi = 5000 \times \pi \times {10^{ - 4}}T{m^2}$
Rate of change of magnetic flux is given by
$\dfrac{{\Delta \phi }}{{\Delta t}} = \dfrac{{0.5\pi }}{4}$
Now calculating emf produced in the coil is given by:
$\varepsilon = - 28 \times \dfrac{{0.5\pi }}{4} \\
\Rightarrow \varepsilon = - 3.5\pi \\
\therefore \varepsilon = - 10.99\,V \\ $
Hence, the emf produced in the coil is $ - 10.99\,V$. The negative sign implies that emf produced in the direction opposite to that of changing flux.

Note: We should know the first law of electromagnetic induction states that whenever a conductor is placed in a varying magnetic field, emf are induced which is called induced emf such that its direction opposes the change in flux inside the conductor. This is also known as Faraday’s first law of electromagnetic induction.