Skip to main content
Back to ELECTRO MAGNETIC INDUCTION

JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Medium
1 of 1
Application +4 | -1
In a transformer, the coefficient of mutual inductance between the primary and the secondary coil is 0.2 henry. When the current changes by 5 ampere/second in the primary, the induced e.m.f. in the secondary will be:
View answer and explanation
Correct option: B
1Β V1 \mathrm{~V}
Option B (1Β V1 \text{ V}) is correct because the induced emf is calculated using the formula e=Mdidte = M \frac{di}{dt}, where M is the mutual inductance and didt\frac{di}{dt} is the rate of change of current.

More Questions from ELECTRO MAGNETIC INDUCTION

Practice All Chapter MCQs β†’
Type at least 2 characters or jump directly:
πŸ”“

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.