The self-inductance of a coil is 2 Henry. If the current in the coil changes at a rate of 5 Amperes per second, what is the induced electromotive force?

C. 10 V — The induced electromotive force in an inductor is given by the product of self-inductance and the rate of change of current. Multiplying 2 Henry by 5 Amperes per second yields 10 Volts.

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$The self-inductance of a coil is L = 2 H . If the current in the coil changes at a rate of \frac{d I}{d t} = 5 A/s, what is the induced electromotive force ε ?$

A. $2.5 V$
B. $7 V$
C. $10 V$
D. $3 V$

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Correct answer: C — $10 V$

Solution

The induced electromotive force is ε = L \frac{d I}{d t} . Substituting the values, ε = 2 \times 5 = 10 V .

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The self-inductance of a coil is L=2 H. If the current in the coil changes at a rate of dtdI​=5 A/s, what is the induced electromotive force ε?

Solution

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More Electromagnetic Induction practice questions

$The self-inductance of a coil is L = 2 H . If the current in the coil changes at a rate of \frac{d I}{d t} = 5 A/s, what is the induced electromotive force ε ?$