Question

A circular coil of radius 10 cm and resistan of $2~\Omega$ is plad with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^\circ$ in 0.25 s. If the magnitude of the indud emf is $3.8 \times 10^{-3} \text{ V}$, then the number of turns of the coil is (Horizontal component of earth's magnetic field at the pla is $3 \times 10^{-5} \text{ T}$)

• A. 504 turns
• B. 458 turns
• C. 302 turns
• D. 608 turns

Hint:

Rotation \& Induced EMF:

• Faraday's Law: $\textemf = -N \fracd\Phidt$
• Magnetic flux: $\Phi = B \cdot A \cdot \cos\theta$
• For $180^\circ$ flip, $\Delta \Phi = 2BA$
• emf = $\frac2NBA\Delta t$

Answer 1

The Correct Option is A

We use Faraday’s law of electromagnetic induction: \[ \text{emf} = \left| \frac{\Delta \Phi}{\Delta t} \right| = \frac{2NBA}{\Delta t} \] The coil rotates through $180^\circ$, causing a change in magnetic flux of $2BA$ per turn. The area of the coil is: \[ A = \pi r^2 = \pi (0.1)^2 = 0.01\pi~\text{m}^2 \] Substituting values: \[ 3.8 \times 10^{-3} = \frac{2 \times N \times 3 \times 10^{-5} \times 0.01\pi}{0.25} \] Solving: \[ N = \frac{3.8 \times 10^{-3} \times 0.25}{2 \times 3 \times 10^{-5} \times 0.01 \pi} \approx 504 \] Hence, the number of turns is 504.

Answer 2

Step 1: Write down the known values
Radius of the coil, \( r = 10 \text{ cm} = 0.1 \text{ m} \)
Resistance, \( R = 2 \, \Omega \) (though not directly needed here)
Time of rotation, \( t = 0.25 \text{ s} \)
Induced emf, \( \mathcal{E} = 3.8 \times 10^{-3} \text{ V} \)
Horizontal component of Earth's magnetic field, \( B = 3 \times 10^{-5} \text{ T} \)

Step 2: Calculate the area of the coil
\[ A = \pi r^2 = \pi \times (0.1)^2 = \pi \times 0.01 = 0.0314 \text{ m}^2 \]

Step 3: Understand the change in magnetic flux
The coil is rotated by \( 180^\circ \) (half a turn) with its plane perpendicular to the magnetic field.
Initial magnetic flux, \( \Phi_i = B A \cos 0^\circ = B A \)
Final magnetic flux, \( \Phi_f = B A \cos 180^\circ = - B A \)
Change in flux, \( \Delta \Phi = \Phi_f - \Phi_i = - B A - B A = -2 B A \)

Step 4: Use Faraday’s law of electromagnetic induction
Induced emf is given by:
\[ \mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N \frac{2 B A}{t} \]
where \( N \) is the number of turns.

Step 5: Calculate the number of turns \( N \)
Rearranging:
\[ N = \frac{\mathcal{E} t}{2 B A} \]
Substitute the known values:
\[ N = \frac{3.8 \times 10^{-3} \times 0.25}{2 \times 3 \times 10^{-5} \times 0.0314} = \frac{9.5 \times 10^{-4}}{1.884 \times 10^{-6}} \approx 504 \]

Step 6: Conclusion
The number of turns in the coil is approximately 504.