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Electromagnetic Induction

JEE Main level, numerical answer, medium

32 of 153

Question 32 of 153

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

Show answer and solution

Answer: 5

I=\left(5{t}^{2}+2t+C\right)

\dfrac{di}{dt}=(10t+2)

{\phi }_{\text{small}}=BA=\left(\dfrac{{\mu }_{0}I{N}_{2}}{2R}\right)\left(\pi {r}^{2}\right)

induced emf in small coil

e=\dfrac{d\phi }{dt}=\left(\dfrac{{\mu }_{0}{N}_{2}}{2r}\right)\pi {r}^{2}{N}_{1}\dfrac{di}{dt}=\left(\dfrac{{\mu }_{0}{N}_{1}{N}_{2}\pi {r}^{2}}{2R}\right)(10t-2)

t=1

e=\left(\dfrac{{\mu }_{0}{N}_{1}{N}_{2}\pi {r}^{2}}{2R}\right)8=4\dfrac{{\mu }_{0}{N}_{1}{N}_{2}\pi {r}^{2}}{R}=\dfrac{4(4\pi ){10}^{-7}\times 200}{20}\times 500\times \dfrac{{10}^{-4}}{{10}^{-2}}\pi

=80\times {\pi }^{2}\times {10}^{-7}\times 10\times {10}^{2}\times {10}^{-2}=8\times {10}^{-4}volt=0.8mV=\dfrac{4}{x}

x=5

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JEE Main 2025 Apr 2 Shift 1 question paper

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