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Step-by-Step Solution
Step 1: Identify the given quantities
• Number of turns, $N = 800$
• Effective area of the coil, $A = 0.05 \text{ m}^2$
• Magnetic field, $B = 5 \times 10^{-5} \text{ T}$
• Time interval for rotation, $\Delta t = 0.1 \text{ s}$
• The coil is rotated by $90^\circ$ (from perpendicular to parallel alignment with respect to the magnetic field).
Step 2: Write down the flux change
When the plane of the coil is perpendicular to the magnetic field, the magnetic flux is:
$\phi_i = BA \quad$ (since the angle between the area vector and $B$ is $0^\circ$).
After rotating by $90^\circ$, the plane of the coil becomes parallel to the magnetic field, so:
$\phi_f = 0 \quad$ (since the coil's area vector is now at $90^\circ$ to $B$, and $\cos 90^\circ = 0$).
Thus, the change in flux $\Delta \phi = \phi_f - \phi_i = 0 - BA = -BA.$
Step 3: Calculate the magnitude of change in flux for all turns
Taking into account the $N$ turns, the total change in flux is
$\Delta \Phi = N \times \Delta \phi = -NBA.$
Numerically:
$ NBA = 800 \times 5 \times 10^{-5} \times 0.05. $
First calculate $ B \times A $:
$ B \times A = (5 \times 10^{-5}) \times 0.05 = 5 \times 0.05 \times 10^{-5} = 0.25 \times 10^{-5} = 2.5 \times 10^{-6}. $
Then multiply by $N = 800$:
$ NBA = 800 \times 2.5 \times 10^{-6} = 2.0 \times 10^{-3}. $
Step 4: Apply Faraday’s law of electromagnetic induction
According to Faraday’s law, the induced emf is
$ e = - \frac{\Delta \Phi}{\Delta t} = - \frac{\phi_f - \phi_i}{\Delta t}.$
Since $\Delta \Phi = -NBA$, its magnitude is $NBA$, so the magnitude of induced emf becomes
$ e = \frac{NBA}{\Delta t}. $
Substitute the values:
$ e = \frac{2.0 \times 10^{-3}}{0.1} = 2.0 \times 10^{-2} = 0.02 \text{ V}. $
Final Answer
The induced emf in the coil is $0.02 \text{ V}$.
