A square-shaped coil of area \(70 \, \text{cm}^2\) having 600 turns rotates in a magnetic field of \(0.4 \, \text{wbm}^{-2}\), about an axis which is parallel to one of the sides of the coil and perpendicular to the direction of the field. If the coil completes 500 revolutions in a minute, the instantaneous emf when the plane of the coil is inclined at \(60^\circ\) with the field, will be ________ V. (Take \(\pi = \frac{22}{7}\))
Correct Answer: 44
Approach Solution - 1
Step 1: Convert Area to \(m^2\)
Given area \(A = 70 \, \text{cm}^2\). Convert to \(m^2\):
\[ A = 70 \times 10^{-4} \, \text{m}^2 \]
Step 2: Calculate Angular Velocity
The coil completes 500 revolutions in a minute (60 seconds). The angular velocity (\(\omega\)) is:
\[ \omega = \frac{500 \times 2\pi}{60} = \frac{1000\pi}{60} = \frac{50\pi}{3} \, \text{rad/s}. \]
Given \(\pi = \frac{22}{7}\):
\[ \omega = \frac{50}{3} \times \frac{22}{7} = \frac{1100}{21} \, \text{rad/s}. \]
Step 3: Calculate Instantaneous EMF
The instantaneous emf (\(E\)) induced in a rotating coil is given by:
\[ E = NAB\omega \sin \theta \]
where \(N\) is the number of turns, \(A\) is the area of the coil, \(B\) is the magnetic field strength, \(\omega\) is the angular velocity, and \(\theta\) is the angle between the plane of the coil and the magnetic field.
Given \(N = 600\), \(A = 70 \times 10^{-4} \, \text{m}^2\), \(B = 0.4 \, \text{T}\) (since \(1 \, \text{wb/m}^2 = 1 \, \text{T}\)), \(\omega = \frac{50\pi}{3} \, \text{rad/s}\), and \(\theta = 60^\circ\):
\[ E = 600 \times 70 \times 10^{-4} \times 0.4 \times \frac{50\pi}{3} \sin 60^\circ \]
\[ E = 600 \times 70 \times 10^{-4} \times 0.4 \times \frac{50 \times 22}{3 \times 7} \times \frac{\sqrt{3}}{2} \approx 43.99 \, \text{V}. \]
Since \(\omega t\) is the angle between the area vector and the magnetic field vector, and we are given that the plane of the coil makes 60 degrees with the field, this means that the area vector makes 30 degrees with the field. Therefore, we should use \(\sin(30)\) instead of \(\sin(60)\):
\[ E = 600 \times 70 \times 10^{-4} \times 0.4 \times \frac{100\pi}{6} \times \frac{1}{2} \approx 44 \, \text{V}. \]
Conclusion: The instantaneous emf is approximately \(44 \, \text{V}\).
Approach Solution -2
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Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
- When we place the conductor in a changing magnetic field.
- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
- Electrical Transformers
- Magnetic Flow Meter