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

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(D) The induced electromotive force $(e)$ in a coil can be expressed in two ways:$1$. Due to the change in magnetic flux linked with the area: $e = B

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$10$

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(D) The induced electromotive force $(e)$ in a coil can be expressed in two ways:$1$. Due to the change in magnetic flux linked with the area: $e = B \cdot \frac{dA}{dt}$$2$. Due to self-inductance: $e = L \cdot \frac{di}{dt}$Equating these two expressions: $B \cdot \frac{dA}{dt} = L \cdot \frac{di}{dt}$Given:$B = 1 \, T$$\frac{dA}{dt} = 5 \, m^2/ms = 5 	imes 10^3 \, m^2/s$$\frac{di}{dt} = \frac{2 - 1}{2 	imes 10^{-3}} = \frac{1}{2 	imes 10^{-3}} = 500 \, A/s$Substituting the values:$1 	imes (5 	imes 10^3) = L 	imes 500$$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 magnetic field is $1 \, T$,then the self-inductance of the coil is......$H$

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