The coefficient of mutual induction between the primary and secondary coil of a transformer is $0.4\text{ H}$. When the current in the primary coil changes at the rate of $10\text{ As}^{-1}$, then the induced emf in the secondary will be:
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Mutual induction relates the voltage in one coil to the rate of change of current in another.
The formula is a direct multiplication: $V = M \times (\text{rate of change of current})$.
Step 1: Check the units before calculating anything. Mutual inductance is measured in henries, and $1\text{ H} = 1\text{ V s A}^{-1}$, so multiplying $M$ (in H) by $\frac{dI}{dt}$ (in A/s) directly gives volts, confirming the formula $e_s = M\dfrac{dI_p}{dt}$ needs no extra conversion. Step 2: Substitute the given values. \[ e_s = (0.4\text{ H})\times(10\text{ A s}^{-1}) \] Step 3: Multiply out. \[ \boxed{e_s = 4.0\text{ V}} \]