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: Understanding the Concept:
By Faraday's law the emf induced in a coil of N turns equals N times the rate of change of flux through one turn.

Step 2: Key Formula or Approach:
\(\varepsilon = N \frac{A\,|\Delta B|}{\Delta t}\), because the plane is perpendicular to the field and the flux per turn is \(BA\).

Step 3: Find the area:
Radius is \(r = 0.01\ \text{m}\). So \(A = \pi r^2 = \pi \times 10^{-4} \approx 3.14 \times 10^{-4}\ \text{m}^2\).

Step 4: Find the change in field:
\(|\Delta B| = 6000 - 1000 = 5000\ \text{T}\), and \(\Delta t = 4\ \text{s}\). So \(\frac{\Delta B}{\Delta t} = 1250\ \text{T/s}\).

Step 5: Solve for N:
\[ N = \frac{\varepsilon \, \Delta t}{A\,\Delta B} = \frac{11 \times 4}{3.14\times 10^{-4} \times 5000} = \frac{11}{0.3927} \approx 28 \]

Step 6: Check the options:
With N = 16, 18 or 24 the emf would be 6.3 V, 7.1 V or 9.4 V, so none matches 11 V. N = 28 gives \(28 \times 0.3927 = 11.0\) V.

Final Answer:
The coil has 28 turns. \[ \boxed{N = 28} \]