Verified Solution • 2025–26 Session moderate

QJEEM20Ph328d: JEE Main 2020 Physics - Class 12 JEEM20Ph328d Solution: Two concentric circular coils, C 1 and C 2 are placed in the XY …

Two concentric circular coils, C1 and C2 are
placed in the XY plane. C1 has 500 turns, and
a radius of 1 cm. C2 has 200 turns and radius
of 20 cm. C2 carries a time dependent current
I(t) = (5t2 – 2t + 3) A where t is in s. The emf
induced in C1 (in mV), at the instant t = 1 s is
4x{4 \over x} . The value of x is ___ .

Step-by-Step Verified Solution

B2=μ0I2N22R2{B_2} = {{{\mu _0}{I_2}{N_2}} \over {2{R_2}}}

Total flux ϕ=N1B2πR12=N1N2μ0I2R2πR12\phi = {N_1}{B_2}\pi {R_1}^2 = {N_1}{N_2}{{{\mu _0}I} \over {2{R_2}}}\pi {R_1}^2

∴\therefore ϕ=500×200×4π×10−7×(5t2−2t−3)π(10−2)22×20×10−2\phi = {{500 \times 200 \times 4\pi \times {{10}^{ - 7}} \times (5{t^2} - 2t - 3)\pi {{({{10}^{ - 2}})}^2}} \over {2 \times 20 \times {{10}^{ - 2}}}}

= 105×4π2×10−7(5t2−2t+3)×10−440×10−2{{{{10}^5} \times 4{\pi ^2} \times {{10}^{ - 7}}(5{t^2} - 2t + 3) \times {{10}^{ - 4}}} \over {40 \times {{10}^{ - 2}}}}

⇒\Rightarrow ϕ=(5t2−2t+3)×10−4\phi = (5{t^2} - 2t + 3) \times {10^{ - 4}}

We know, e=∣dϕdt∣=(10t−2)×10−4e = \left| {{{d\phi } \over {dt}}} \right| = (10t - 2) \times {10^{ - 4}}

At t=1sec⁡t = 1\sec

e=8×10−4=0.8 mV=0.810=45e = 8 \times {10^{ - 4}} = 0.8\,mV = {{0.8} \over {10}} = {4 \over 5}

∴\therefore x=5x = 5

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