Two coils A and B have mutual inductance 2- 10-2 henry- If the current in the primary is i - 5sin 10- t then the maximum value of e-m-f- induced in coil B is-

The correct option is A π volt emf= M df dt = 2× 1 100 × d dt (5sin 10π t )= 2× 1 100 × 5× 10π×cos 10π t = πcos 10π t #160; Maximum value o ...

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Standard X

Physics

Magnetic Field Due to Straight Current Carrying Conductor

Two coils A a...

Question

Two coils A and B have mutual inductance $2 \times 10^{- 2}$ henry. If the current in the primary is $i = 5 sin (10 π t)$ then the maximum value of e.m.f. induced in coil B is:

π v o l t

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\frac{π}{2} v o l t

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\frac{π}{3} v o l t

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\frac{π}{4} v o l t

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Solution

The correct option is A $π v o l t$
$e m f = - M \frac{d f}{d t} = - 2 \times \frac{1}{100} \times \frac{d}{d t} (5 sin 10 π t) = - 2 \times \frac{1}{100} \times 5 \times 10 π \times cos 10 π t = - π cos 10 π t$
Maximum value of emf $= π v o l t$

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Two coils A and B have mutual inductance 2×10−2 henry. If the current in the primary is i=5sin(10πt) then the maximum value of e.m.f. induced in coil B is:

The correct option is A πvoltemf=−Mdfdt=−2×1100×ddt(5sin10πt)=−2×1100×5×10π×cos10πt=−πcos10πtMaximum value of emf=πvolt

Two coils A and B have mutual inductance $2 \times 10^{- 2}$ henry. If the current in the primary is $i = 5 sin (10 π t)$ then the maximum value of e.m.f. induced in coil B is:

The correct option is A $π v o l t$
$e m f = - M \frac{d f}{d t} = - 2 \times \frac{1}{100} \times \frac{d}{d t} (5 sin 10 π t) = - 2 \times \frac{1}{100} \times 5 \times 10 π \times cos 10 π t = - π cos 10 π t$
Maximum value of emf $= π v o l t$