Question

A coil of resistance $16 \Omega$ is placed with its plane perpendicular to a uniform magnetic field whose flux ($\phi$ in $10^{-3}$ weber) changes with time (t in second) as $\phi = 5t^2+4t+2$. The induced current at time $t=6$ seconds is

• A. 4 mA
• B. 2.12 mA
• C. 34 mA
• D. 74 mA

Hint:

Faraday's Law of Induction, $\varepsilon = \left| - \frac{d\phi}{dt} \right|$, and Ohm's Law, $I = \varepsilon/R$, are the two steps for finding induced current. Always ensure all units are in the SI system (Webers, seconds, Volts, Amperes, Ohms).

Answer 1

The Correct Option is D

Step 1: Faraday's Law of Induction
The magnitude of the induced electromotive force (e.m.f.) in a coil with $N$ turns is given by \[ \varepsilon = \left| - \frac{d\phi}{dt} \right|. \] For a single-turn coil ($N=1$), this simplifies to: \[ \varepsilon = \left| \frac{d\phi}{dt} \right|. \]
Step 2: Magnetic Flux and Induced e.m.f.
The magnetic flux through the coil is given as \[ \phi = (5t^2 + 4t + 2) \times 10^{-3} \text{ Wb}. \] The induced e.m.f. is the time derivative of flux: \[ \varepsilon = \frac{d\phi}{dt} = \frac{d}{dt} \Big[(5t^2 + 4t + 2) \times 10^{-3}\Big] = (10t + 4) \times 10^{-3} \text{ V}. \]
Step 3: Induced e.m.f. at $t=6$ s
\[ \varepsilon = (10 \cdot 6 + 4) \times 10^{-3} \text{ V} = 64 \times 10^{-3} \text{ V} = 0.064 \text{ V}. \]
Step 4: Induced Current Using Ohm's Law
The induced current is \[ I = \frac{\varepsilon}{R}. \] The given resistance is $R = 16~\Omega$, so \[ I = \frac{0.064}{16} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}. \]
Step 5: Note on Answer Key Discrepancy
The calculated current is $4~\text{mA}$ (Option A). However, the provided answer key states $74~\text{mA}$ (Option D). This indicates a likely typo in the question's flux value or resistance. To match the keyed answer, the flux or resistance would need to be different, but the physically correct solution based on the given data is \[ \boxed{I = 4~\text{mA}}. \] Conclusion:
Using Faraday's Law and Ohm's Law, the induced current at $t=6$ s is $4~\text{mA}$. The key answer ($74~\text{mA}$) does not match the given problem data.