A current of 2 A passing through a coil of 100 turns gives rise to a magnetic flux of 5×10⁻³ Wb per turn. The magnetic energy associated with the coil is .

Question

Accepted Answer

(4) 0.5 J. Given: I=2 A, N=100, flux per turn Φ=5×10⁻³ Wb. Idea: the flux linkage NΦ equals LI, which gives L; the stored energy is then 1/2LI². L=(NΦ)/I=(100×5×10⁻³)/2=0.25 H. U=1/2LI²=1/2×0.25×2². So U=0.5 J.

A current of 2 A passing through a coil of 100 turns gives rise to a magnetic flux of 5×10⁻³ Wb per turn. The magnetic energy associated with the coil is ______.

Step-by-step solution

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