Question

In an ac generator, a rectangular coil of $100$ turns each having area $14 \times 10^{-2} m ^2$ is rotated at $360\, rev / min$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 T$ The maximum value of the emf produced will be ________ $V$
(Take \(\pi =\frac{22}{7}\))

Answer 1

Correct Answer: 1584

To determine the maximum value of the electromotive force (emf) produced by the ac generator, we use the formula for the maximum emf: \(ε_{max} = NABω\), where:

• N is the number of turns in the coil.
• A is the area of one turn of the coil.
• B is the magnetic field strength.
• ω is the angular velocity in radians per second.

First, convert the given rotational speed from revolutions per minute to radians per second: \(ω = 360 \, \text{rev/min} \times \frac{2\pi \, \text{rad/rev}}{60 \, \text{s/min}} = \frac{360 \times 2 \times\frac{22}{7}}{60} = 37.68 \, \text{rad/s}\).
Now, substitute the given values into the maximum emf formula:
\(ε_{max} = (100 \, \text{turns}) \times (14 \times 10^{-2} \, \text{m}^2) \times (3.0 \, \text{T}) \times (37.68 \, \text{rad/s})\)
Calculating further:
\(ε_{max} = 100 \times 0.14 \times 3 \times 37.68 = 1584 \, \text{V}\).
The calculated maximum emf is 1584 V, which is within the expected range of 1584 to 1584 V.