Question

The magnetic flux \(\phi\) (in weber) linked with a closed circuit of resistance \(8 \, \Omega\) varies with time (in seconds) as \(\phi = 5t^2 - 36t + 1\). The induced current in the circuit at \(t = 2 \, \text{s}\) is ______ A.

Answer 1

Correct Answer: 2

To find the induced current in the circuit at \( t = 2 \, \text{s} \), we first need to determine the electromotive force (emf) induced in the circuit using Faraday's Law of electromagnetic induction. The emf (\( \varepsilon \)) is given by the negative rate of change of magnetic flux (\( \phi \)) with respect to time, \( \varepsilon = -\frac{d\phi}{dt} \). Given \( \phi = 5t^2 - 36t + 1 \), we differentiate: \(\frac{d\phi}{dt} = \frac{d}{dt}(5t^2 - 36t + 1) = 10t - 36\).

Substitute \( t = 2 \) into the derivative:
\(\left.\frac{d\phi}{dt}\right|_{t=2} = 10(2) - 36 = 20 - 36 = -16 \, \text{Wb/s}.\)

The induced emf is \(\varepsilon = -\left(-16\right) = 16 \, \text{V}.\)

Using Ohm's Law, \( I = \frac{\varepsilon}{R} \), with resistance \( R = 8 \, \Omega \):
\( I = \frac{16}{8} = 2 \, \text{A}.\)

The computed current, \( 2 \, \text{A} \), fits the expected range of 2 to 2 A. Therefore, the induced current at \( t = 2 \, \text{s} \) is \( 2 \, \text{A}. \)

Answer 2

The emf \( \varepsilon \) induced in the circuit is given by Faraday’s law:

\[ \varepsilon = -\frac{d\Phi}{dt}. \]

Calculate \( \frac{d\Phi}{dt} \):

\[ \frac{d\Phi}{dt} = 10t - 36. \]

At \( t = 2 \, \text{s} \):

\[ \varepsilon = -(10 \cdot 2 - 36) = -(-16) = 16 \, \text{V}. \]

The induced current \( i \) in the circuit is:

\[ i = \frac{\varepsilon}{R} = \frac{16}{8} = 2 \, \text{A}. \]

Thus, the induced current at \( t = 2 \, \text{s} \) is:

\[ 2 \, \text{A}. \]