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:

• Inductance, \(L = 15 \times 10^{-3} \, H}\)
• Initial current, \(I_{initial}} = 0 \, A}\)
• Final current, \(I_{final}} = 4 \, A}\)
• Time interval, \(\Delta t = 0.004 \, s}\)

Formula: The induced electromotive force (emf) in an inductor is calculated using: \[emf} = -L \frac{\Delta I}{\Delta t}\] {Solution:} 1. Determine the change in current (\(\Delta I\)): \[\Delta I = I_{final}} - I_{initial}} = 4 \, A} - 0 \, A} = 4 \, A}\] 2. Substitute the given values into the emf formula: \[emf} = -L \frac{\Delta I}{\Delta t}\] \[emf} = -(15 \times 10^{-3} \, H}) \cdot \frac{4 \, A}}{0.004 \, s}}\] 3. Simplify the expression: \[emf} = -(15 \times 10^{-3}) \cdot 1000\] \[emf} = -15 \, V}\] The negative sign denotes the direction of the induced emf. The magnitude is required. Final Answer: \[\boxed{15.0 \, V}}\]