Question

A coil with an average diameter of 0.02 m is placed with its plane perpendicular to a magnetic field of 6000 T. The induced emf in the coil is 11 V, when the magnetic field is changed to 1000 T in 4 s. The number of turns in the coil is

• A. 16
• B. 18
• C. 24
• D. 28

Hint:

Use \(\varepsilon = N A\, \Delta B/\Delta t\) with \(A = \pi r^2\) and \(r = 0.01\) m.

Answer 1

The Correct Option is D

Step 1: Flux at the start and end:
Flux through one turn at first is $\Phi_1 = 6000 \times \pi (0.01)^2 = 0.6\pi\ \text{Wb} \approx 1.885\ \text{Wb}$. At the end it is $\Phi_2 = 1000 \times \pi (0.01)^2 = 0.1\pi \approx 0.314\ \text{Wb}$.

Step 2: Change in flux per turn:
$\Delta\Phi = 0.5\pi \approx 1.571\ \text{Wb}$. In 4 s the average emf per turn is $1.571/4 = 0.3927\ \text{V}$.

Step 3: Count the turns:
All turns are in series, so their emfs add. \[ N = \frac{11}{0.3927} = 28.0 \]

Step 4: Conclusion:
The whole number nearest to 28.0 is 28, which is the fourth option.

Final Answer:
The number of turns is 28. \[ \boxed{28} \]