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प्रश्न
A coil of self-inductance 3 H carries a steady current of 2 A. What is the energy stored in the magnetic field of the coil?
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उत्तर
Energy stored in the magnetic field, UB = `1/2 "LI"^2 = 1/2 xx 3 xx 2^2 = 6` J
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संबंधित प्रश्न
Explain the phenomenon of self induction
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(a) magnetic field at the centre
(b) magnetic flux linked with the solenoid
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(d) rate of Joule heating.
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(Assume that, l = current passes through the turns, µ0 = Permeability of vacuum)
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- self-inductance of coil.
- Phase difference between the voltage and the current.
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