Step 1: Understanding the Question:
We must apply Faraday's Law of Electromagnetic Induction to find the time duration ($t$) over which a coil is completely removed from a magnetic field.
Step 2: Detailed Explanation:
Faraday's law states that the magnitude of the average induced EMF ($e$) is equal to the rate of change of magnetic flux ($\Phi$):
$e = \frac{N \cdot \Delta \Phi}{\Delta t}$
The magnetic flux is given by $\Phi = B \cdot A \cdot \cos \theta$.
Since the field acts "at right angles to the coil," the angle between the magnetic field vector and the area vector (the normal to the coil) is $0^\circ$, so $\cos(0^\circ) = 1$.
Initial Flux:
$B_{initial} = 4 \times 10^{-2} \text{ T}$
Area $A = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 = 10^{-2} \text{ m}^2$
$\Phi_{initial} = B_{initial} \times A = (4 \times 10^{-2}) \times (10^{-2}) = 4 \times 10^{-4} \text{ Wb}$
Final Flux:
The coil is completely removed from the field, so $B_{final} = 0 \text{ T}$.
$\Phi_{final} = 0 \text{ Wb}$
Change in Flux ($\Delta \Phi$):
Magnitude of change $|\Delta \Phi| = 4 \times 10^{-4} \text{ Wb}$.
We are given:
Number of turns ($N$) = 50
Induced EMF ($e$) = 0.1 V
Substitute these values into Faraday's equation:
$e = \frac{N \cdot |\Delta \Phi|}{t}$
$0.1 = \frac{50 \times (4 \times 10^{-4})}{t}$
Simplify the numerator:
$50 \times 4 = 200$
$0.1 = \frac{200 \times 10^{-4}}{t}$
$0.1 = \frac{2 \times 10^{-2}}{t}$
Rearrange to solve for $t$:
$t = \frac{0.02}{0.1}$
$t = 0.2 \text{ seconds}$
Step 3: Final Answer:
The value of 't' is 0.2 second, matching option (c).