Question

The current in a coil of \(15 \, mH}\) increases uniformly from zero to \(4 \, A}\) in \(0.004 \, s}\). Calculate the electromotive force (emf) induced in the coil:

• A. \(22.5 \, V}\)
• B. \(17.5 \, V}\)
• C. \(15.0 \, V}\)
• D. \(12.5 \, V}\)

Hint:

When dealing with induction problems, always remember that the sign of the induced emf depends on the direction of the change in current and the polarity of the inductor but is usually reported as a magnitude in problems unless specifically dealing with circuit directions.

Answer 1

The Correct Option is C

Given: \begin{itemize} \item Inductance, \(L = 15 \, mH} = 15 \times 10^{-3} \, H}\)
\item Initial current, \(I_{initial}} = 0 \, A}\)
\item Final current, \(I_{final}} = 4 \, A}\)
\item Time, \(\Delta t = 0.004 \, s}\)
\end{itemize} Formula:
The induced emf in an inductor is given by:
\[ emf} = -L \frac{\Delta I}{\Delta t} \] {Solution:} 1. Calculate the change in current (\(\Delta I\)):
\[ \Delta I = I_{final}} - I_{initial}} = 4 \, A} - 0 \, A} = 4 \, A} \] 2. Substitute the values into the formula:
\[ emf} = -L \frac{\Delta I}{\Delta t} \] \[ emf} = -(15 \times 10^{-3} \, H}) \cdot \frac{4 \, A}}{0.004 \, s}} \] 3. Simplify the calculation:
\[ emf} = -(15 \times 10^{-3}) \cdot 1000 \] \[ emf} = -15 \, V} \] The negative sign indicates the direction of the induced emf, but we are only interested in the magnitude.
Final Answer: \[ \boxed{15.0 \, V}} \]