The current in an inductor of self-inductance $4 ext{ H}$ changes from $4 ext{ A}$ to $2 ext{ A}$ in $1

Question

The current in an inductor of self-inductance $4 	ext{ H}$ changes from $4 	ext{ A}$ to $2 	ext{ A}$ in $1 	ext{ s}$. The emf induced in the coil is:

Accepted Answer

The induced electromotive force (emf) $\mathcal{E}$ in an inductor is given by Faraday's law of induction: $\mathcal{E} = -L \frac{dI}{dt}$.

1.  **Identify Given Values:**
    *   Self-inductance ($L$) = $4 	ext{ H}$
    *   Initial Current ($I_i$) = $4 	ext{ A}$
    *   Final Current ($I_f$) = $2 	ext{ A}$
    *   Time interval ($\Delta t$) = $1 	ext{ s}$

2.  **Calculate Rate of Change of Current:**
    $\frac{dI}{dt} = \frac{I_f - I_i}{\Delta t} = \frac{2 - 4}{1} = -2 	ext{ A/s}$

3.  **Calculate Induced EMF:**
    $\mathcal{E} = - (4 	ext{ H}) 	imes (-2 	ext{ A/s})$
    $\mathcal{E} = 8 	ext{ V}$.

The positive sign indicates the direction opposes the change in current (Lenz's Law), but the magnitude is 8 V.

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