Question

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 \(\pi = \frac{22}{7}\))

Answer 1

Correct Answer: 44

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 \(44 \, \text{V}\).

Answer 2

The correct answer is 44.
$N=600,A=70\times 10^{-4}{m}^2,B=0.4T$
$\omega =\frac{500\times 2\pi }{60}=\frac{100\pi }{6}rad/s$
$E=NAB\omega \sin \omega t$
$\omega t\text{is angle}b/w\vec{A}\&\vec{B}$
$=600\times 70\times 10^{-4}\times 0.4\times \frac{100\pi }{6}\times \frac{1}{2}$
$=44V$