If in a coil,the rate of change of area is 5 | vedclass

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(D) The magnetic flux linked with the coil is $\phi = B \cdot A$. For a coil with $N$ turns,the flux linkage is $N\phi = Li$,where $L$ is the self-ind

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$10$

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(D) The magnetic flux linked with the coil is $\phi = B \cdot A$. For a coil with $N$ turns,the flux linkage is $N\phi = Li$,where $L$ is the self-inductance. Differentiating with respect to time $t$,we get: $N \frac{d\phi}{dt} = L \frac{di}{dt}$. Since $\phi = BA$,we have $NB \frac{dA}{dt} = L \frac{di}{dt}$. Given: $\frac{dA}{dt} = 5 \, m^2/ms = 5 	imes 10^3 \, m^2/s$,$B = 1 \, T$,$di = (2 - 1) \, A = 1 \, A$,$dt = 2 	imes 10^{-3} \, s$,and assuming $N = 1$. Substituting the values: $1 	imes 1 	imes (5 	imes 10^3) = L 	imes \left( \frac{1}{2 	imes 10^{-3}} ight)$. $5000 = L 	imes 500$. $L = \frac{5000}{500} = 10 \, H$.

If in a coil,the rate of change of area is $5 \, m^2/ms$ and the current changes from $2 \, A$ to $1 \, A$ in $2 \times 10^{-3} \, s$. If the magnitude of the magnetic field is $1 \, T$,then the self-inductance of the coil is ... $H$.

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