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 cm, $C 2$ has 200 turns and radius of 20cm. $C 2$ carries a time dependent current $I (t) = 5 t 2 − 2 t + 3 A$ where t is in s. The emf induced in $C 1 (in mV)$ at the instant $t = 1 s is 4 x$ . The value of x is
$[Take π^{2} = 10]$

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answer is 5.

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Detailed Solution

For coil $C_{1}$ , No of turns $N 1 = 500$ and radius $r = 1 cm$

For coil $C 2$ , No of turns $N 2 = 200$ and radius $r = 20 cm$

$I = (5 t^{2} - 2 t + 3)$

$\Rightarrow \frac{d I}{d t} = (10 t - 2)$

$ϕ_{small} = B A = (\frac{μ_{0} I N_{0}}{2 R}) (π r^{2})$
Induced emf in small coil , $e = \frac{d ϕ}{d t} = (\frac{μ_{0} N_{2}}{2 R}) π r^{2} N_{1} \frac{d I}{d t} = (\frac{μ_{0} N_{1} N_{2} π r^{2}}{2 R}) (10 t - 2)$
At $t = 1 s$

$e = (\frac{μ_{0} N_{1} N_{2} π r^{2}}{2 R}) 8 = 4 \frac{μ_{0} N_{1} N_{2} π r^{2}}{R}$

$\Rightarrow e = \frac{4 (4 π) 10^{- 7} \times 200}{20} \times 500 \times \frac{10^{- 4}}{10^{- 2}} π$
$= 8 \times 10^{- 4} volt = 0.8 mV = \frac{4}{x}$

$\Rightarrow x = 5$

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