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:

• A. $\quad \frac{2\pi V}{5} \\$
• B. $\quad \frac{\pi V}{4} \\$
• C. $\quad \frac{4\pi V}{5} \\$
• D. $\quad \frac{12\pi V}{5}$

Answer 1

The Correct Option is C

The maximum induced EMF in a rotating coil is given by the formula:
ε_max = 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.

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

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

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

Answer 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:

ε_max = NBAω

Substituting the values:

ε_max = 200 × 0.02 × 0.1 × 2π

ε_max = 200 × 0.02 × 0.1 × 2π = 0.8π V

ε_max = $\frac{4}{5}\pi$ 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).