Question

A rectangular coil of 250 turns has an average area of \( 20 \, cm \times 15 \, cm \). The coil rotates with a speed of 60 cycles per second in a uniform magnetic field of \( 2 \times 10^{-2} \, T \) about an axis perpendicular to the field. The peak value of the induced emf is:

• A. \( 36\pi \) volt
• B. \( 30\pi \) volt
• C. \( 18\pi \) volt
• D. \( 12\pi \) volt

Hint:

For a rotating coil, peak emf is \( E_0 = NBA\omega \). Always convert area from \( cm^2 \) to \( m^2 \) before substitution.

Answer 1

The Correct Option is C

Step 1: Write the formula for peak induced emf.
For a rotating coil in a uniform magnetic field, peak induced emf is:
\[ E_0 = NBA\omega \]
where \( N \) is number of turns, \( B \) is magnetic field, \( A \) is area of coil, and \( \omega \) is angular velocity.

Step 2: Write the given values.
\[ N = 250 \]
\[ B = 2 \times 10^{-2} \, T \]
\[ f = 60 \, s^{-1} \]

Step 3: Convert area into SI unit.
Area of rectangular coil is:
\[ A = 20 \, cm \times 15 \, cm \]
\[ A = 300 \, cm^2 \]
Since \( 1 \, cm^2 = 10^{-4} \, m^2 \),
\[ A = 300 \times 10^{-4} = 3 \times 10^{-2} \, m^2 \]

Step 4: Calculate angular velocity.
\[ \omega = 2\pi f \]
\[ \omega = 2\pi \times 60 \]
\[ \omega = 120\pi \, rad \, s^{-1} \]

Step 5: Substitute values in formula.
\[ E_0 = NBA\omega \]
\[ E_0 = 250 \times 2 \times 10^{-2} \times 3 \times 10^{-2} \times 120\pi \]

Step 6: Simplify the expression.
\[ 250 \times 2 \times 10^{-2} = 5 \]
\[ 5 \times 3 \times 10^{-2} = 0.15 \]
\[ E_0 = 0.15 \times 120\pi \]
\[ E_0 = 18\pi \, V \]

Step 7: Conclusion.
Thus, the peak value of induced emf is:
\[ \boxed{18\pi \, volt} \]