Question

An emf is generated by an ac generator having 100 turn coil, of loop area 1 m². The coil rotates at a speed of one revolution per second and placed in a uniform magnetic field of 0.05 T perpendicular to the axis of rotation of the coil. The maximum value of emf is :-

• A. 3.14 V
• B. 31.4 V
• C. 62.8 V
• D. 6.28 V

Answer 1

Answer: (B)

31.4 V

Answer 2

The maximum value of emf generated by an ac generator is given by the formula:

$\mathcal{E}_{max} = NBA\omega$

Where:

$N$ = Number of turns in the coil $B$ = Magnetic field strength $A$ = Area of the coil $\omega$ = Angular speed of rotation

Given values:

Number of turns, $N = 100$ Loop area, $A = 1 \text{ m}^2$ Magnetic field, $B = 0.05 \text{ T}$ Speed of rotation, $f = 1 \text{ revolution per second}$

First, calculate the angular speed $\omega$ from the frequency $f$:

$\omega = 2\pi f$

$\omega = 2\pi (1 \text{ rev/s})$

$\omega = 2\pi \text{ rad/s}$

Now, substitute all the values into the formula for $\mathcal{E}_{max}$:

$\mathcal{E}_{max} = (100) \times (0.05 \text{ T}) \times (1 \text{ m}^2) \times (2\pi \text{ rad/s})$

$\mathcal{E}_{max} = 5 \times 2\pi$

$\mathcal{E}_{max} = 10\pi \text{ V}$

Using the approximate value of $\pi \approx 3.14$:

$\mathcal{E}_{max} = 10 \times 3.14$

$\mathcal{E}_{max} = 31.4 \text{ V}$

The maximum value of emf is 31.4 V.