A 50 mH coil carries a current of 2A. The energy stored in its magnetic field is:
Correct answer: C. E = 0.1 J
- A. E = 005 J
- B. E = 10J
- C. E = 0.1 J
- D. E = 50 J
Explanation
To calculate the energy stored in the magnetic field of the coil, we can use the formula: E=1/2 LI2where:E is the energy stored in the magnetic field (in joules),L is the inductance of the coil (in Henries), andI is the current flowing through the coil (in amperes).Given that the inductance (L) of the coil is 50 mH=50×10−3 H and the current (I) is 2 A, we can substitute these values into the formula:E=1/2×50×10−3×(2)2E=1/2 ×50×10−3×4E=1/2×0.05×4E=0.1JTherefore, the energy stored in the magnetic field of the coil is 0.1 J0.1J. So, the correct option is: E=0.1J
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About Electromagnetic Induction
Changing magnetic flux induces an emf according to Faraday's law, while Lenz's law gives the direction that opposes the change producing it. Transformers apply induction between coils to step alternating voltage up or down, with turns ratio, current transformation and energy losses determining practical operation.
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