Exams › NEET › Physics

What is the self-inductance of a coil which produces 5V when the current changes from 3 ampere to 2 ampere in one millisecond?

1. 5000 henry
2. 5 milli-henry
3. 50 henry
4. 5 henry

Correct answer: 5 milli-henry

Solution

The self-inductance is calculated using the formula E = -L (dI/dt). Here, E = 5 V, dI = 3 - 2 = 1 A, and dt = 1 ms = 10^-3 s. Substituting, L = E / (dI/dt) = 5 / (1 / 10^-3) = 5 × 10^-3 H = 5 mH.

Related NEET Physics questions

• Which of the following statement regarding transformer is incorrect?
• Which one of the following statements is true?
• Which statement is incorrect related to induced electric field due to changing magnetic flux?
• A dynamo produces an electric current. It is based on the principle:
• In electromagnetic induction, the induced E.M.F is independent of
• A conductor is moved in a varying magnetic field. Name the law which determines the direction of current induced in the conductor:

⚔️ Practice NEET Physics free + battle 1v1 →