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
$[Take π^{2} = 10]$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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 $t = 1 s$

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

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