Question:

A 200-turn circular coil of area 103cm2 rotates at 60 revolutions per minute in a uniform magnetic field of 0.02 T perpendicular to the axis of rotation of the coil. The maximum voltage induced in the coil is:

Updated On: May 27, 2025
  • \(\quad \frac{2\pi V}{5} \\\)
  • \(\quad \frac{\pi V}{4} \\\)
  • \(\quad \frac{4\pi V}{5} \\\)
  • \(\quad \frac{12\pi V}{5}\)
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The Correct Option is C

Approach Solution - 1

The maximum induced EMF in a rotating coil is given by the formula: 
εmax = NABω
where:- N is the number of turns of the coil, A is the area of the coil in square meters, B is the magnetic field in Tesla, ω is the angular velocity in radians per second.
 

\(\text{Given values: } N = 200, \, A = 10^3 \, \text{cm}^2 = 10^{-1} \, \text{m}^2, \, B = 0.02 \, \text{T}, \, \omega = 2\pi \times\) \(\frac{60}{60} = 2\pi \, \text{rad/s}.\\\)
\(\text{Substituting the values into the formula:}\)

\[\varepsilon_{\text{max}} = 200 \times 10^{-1} \times 0.02 \times 2\pi = \frac{4\pi V}{5}\]

\(\text{Thus, the maximum voltage induced in the coil is } \frac{4\pi V}{5}, \text{ which corresponds to Option (3).}\)

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Approach Solution -2

Given:

  • Number of turns (N) = 200
  • Area (A) = 103 cm² = 0.1 m² (since 104 cm² = 1 m²)
  • Angular velocity (ω) = 60 rpm = 2π rad/s (since ω = 2π × rpm/60)
  • Magnetic field (B) = 0.02 T

The maximum induced voltage is given by:

εmax = NBAω

Substituting the values:

εmax = 200 × 0.02 × 0.1 × 2π

εmax = 200 × 0.02 × 0.1 × 2π = 0.8π V

εmax = \(\frac{4}{5}\pi\) V

The maximum voltage induced in the coil is:

\(\frac{4}{5}\pi \, \text{V}\)

Matching with the given options, the correct answer is option (3).

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