If a current of 4 A produces a magnetic flux | vedclass

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(B) Given: Current $I = 4 \ A$,Number of turns $N = 400$,Magnetic flux $\phi = 3 	imes 10^{-3} \ Wb$.The magnetic flux linkage is given by $N\phi = L

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$2.4$

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(B) Given: Current $I = 4 \ A$,Number of turns $N = 400$,Magnetic flux $\phi = 3 	imes 10^{-3} \ Wb$.The magnetic flux linkage is given by $N\phi = LI$,where $L$ is the self-inductance of the coil.First,calculate the self-inductance $L$:$L = \frac{N\phi}{I} = \frac{400 	imes 3 	imes 10^{-3}}{4} = 100 	imes 3 	imes 10^{-3} = 0.3 \ H$.The energy stored in an inductor is given by the formula $U = \frac{1}{2} LI^2$.Substituting the values:$U = \frac{1}{2} 	imes 0.3 	imes (4)^2$$U = \frac{1}{2} 	imes 0.3 	imes 16$$U = 0.3 	imes 8 = 2.4 \ J$.

If a current of $4 \ A$ produces a magnetic flux of $3 \times 10^{-3} \ Wb$ through a coil of $400$ turns,the energy stored in the coil will be: (in $J$)

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