JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 3)

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
$${4 \over x}$$. The value of x is ___ .
Answer
5

Explanation

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

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

$$ \therefore $$ $$\phi = {{500 \times 200 \times 4\pi \times {{10}^{ - 7}} \times (5{t^2} - 2t - 3)\pi {{({{10}^{ - 2}})}^2}} \over {2 \times 20 \times {{10}^{ - 2}}}}$$

= $${{{{10}^5} \times 4{\pi ^2} \times {{10}^{ - 7}}(5{t^2} - 2t + 3) \times {{10}^{ - 4}}} \over {40 \times {{10}^{ - 2}}}}$$

$$ \Rightarrow $$ $$\phi = (5{t^2} - 2t + 3) \times {10^{ - 4}}$$

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

At $$t = 1\sec $$

$$e = 8 \times {10^{ - 4}} = 0.8\,mV = {{0.8} \over {10}} = {4 \over 5}$$

$$ \therefore $$ $$x = 5$$

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