Question

In a current carrying coil of inductance 60 mH,the current is changed from 2.5 A in one direction to 2.5 A in the opposite direction in 0.10 sec. The average induced EMF in the coil will be:

• A. 1.2 V
• B. 2.4 V
• C. 3.0 V
• D. 1.8 V
• E. 0.6 V

Answer 1

The Correct Option is C

Given parameters:

• Inductance: \( L = 60 \, \text{mH} = 60 \times 10^{-3} \, \text{H} \)
• Current change: \( ΔI = 2.5 - (-2.5) = 5.0 \, \text{A} \)
• Time interval: \( Δt = 0.10 \, \text{s} \)

 

Induced EMF calculation: \[ \text{EMF} = -L \frac{ΔI}{Δt} = -60 \times 10^{-3} \times \frac{5.0}{0.10} \] \[ \text{EMF} = -3.0 \, \text{V} \] (Magnitude: 3.0 V)

Thus, the correct option is (C): 3.0 V.

Answer 2

To find the average induced EMF in a coil when the current changes, we can use Faraday's law of electromagnetic induction, which states that the induced EMF (\(\mathcal{E}\)) in a coil is given by:

\[\mathcal{E} = -L \frac{\Delta I}{\Delta t}\]

where:
- \(L\) is the inductance of the coil,
- \(\Delta I\) is the change in current,
- \(\Delta t\) is the time over which the change occurs.

Given:
- Inductance, \(L = 60 \text{ mH} = 60 \times 10^{-3} \text{ H}\)
- Initial current, \(I_i = 2.5 \text{ A}\)
- Final current, \(I_f = -2.5 \text{ A}\)
- Time interval, \(\Delta t = 0.10 \text{ s}\)

First, calculate the change in current (\(\Delta I\)):
\[\Delta I = I_f - I_i = (-2.5 \text{ A}) - (2.5 \text{ A}) = -2.5 \text{ A} - 2.5 \text{ A} = -5 \text{ A}\]

Now, substitute the values into Faraday's law equation:
\[\mathcal{E} = -L \frac{\Delta I}{\Delta t} = - (60 \times 10^{-3} \text{ H}) \frac{-5 \text{ A}}{0.10 \text{ s}}\]

Simplify the expression:
\[\mathcal{E} = - (60 \times 10^{-3}) \times \frac{-5}{0.10}\]
\[\mathcal{E} = (60 \times 10^{-3}) \times 50\]
\[\mathcal{E} = 3.0 \text{ V}\]
Thus The correct answer is Option (C):\(3.0 V\)