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PhysicsEMI/ACInduced Electric FieldEasy2 minPYQ_2017

PhysicsEasysingle choice

The current in a coil of $L = 40 mH$ is to be increased uniformly from 1 A to $11 A$ in 4 milli sec. The induced e.m.f. will be

Answer:

Solution:

$e = \frac{LdI}{dt} = \frac{40 \times 1 0 ^{- 3} ( 11 - 1 )}{4 \times 1 0 ^{- 3}} = 100 V$

Stream:VITEEESubject:PhysicsTopic:EMI/ACSubtopic:Induced Electric Field

⏱ 2mℹ️ Source: PYQ_2017

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