JEE Main 2020 — Physics Electromagnetism

2020integerhard

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

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

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Solution

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

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

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

induced emf in small coil

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

at $t=1$

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

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

$x=5$

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JEE Main Electromagnetism in other years

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JEE Main 2020 — Physics Electromagnetism

2020integerhard

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Did you get this right?

Solution

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

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

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

induced emf in small coil

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

at $t=1$

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

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

$x=5$

Your note

JEE Main Electromagnetism in other years

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JEE Main 2020 — Physics Electromagnetism

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Solution

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JEE Main 2020 — Physics Electromagnetism

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism