Q176 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
A rectangular coil of 250 turns has an average area of $20 \mathrm{~cm} \times 15 \mathrm{~cm}$. The coil rotates with a speed of 60 cycles per second in a uniform magnetic field of $2 \times 10^{-2} \mathrm{~T}$ about an axis perpendicular to the field. The peak value of the induced emf is:
- A$36 \pi$ volt
- Correct answerB$18 \pi$ volt
- C$12 \pi$ volt
- D$30 \pi$ volt
Solution
The peak value of the induced electromotive force (emf) in a rotating coil is given by the formula $\varepsilon_{0} = N A B \omega$, where $N$ is the number of turns, $A$ is the area of the coil, $B$ is the magnetic field strength, and $\omega$ is the angular velocity.
Given values:
$N = 250$
$A = 20 \text{ cm} \times 15 \text{ cm} = 0.2 \text{ m} \times 0.15 \text{ m} = 0.03 \text{ m}^{2}$
$B = 2 \times 10^{-2} \text{ T}$
Frequency $f = 60 \text{ Hz}$
Angular velocity $\omega = 2 \pi f = 2 \pi \times 60 = 120 \pi \text{ rad/s}$
Substituting these values into the formula:
$\varepsilon_{0} = 250 \times 0.03 \times (2 \times 10^{-2}) \times 120 \pi$
Calculating the product:
$\varepsilon_{0} = 250 \times 0.03 \times 0.02 \times 120 \pi$
$\varepsilon_{0} = 7.5 \times 0.02 \times 120 \pi$
$\varepsilon_{0} = 0.15 \times 120 \pi$
$\varepsilon_{0} = 18 \pi \text{ volt}$
Answer: $18 \pi$ volt
Given values:
$N = 250$
$A = 20 \text{ cm} \times 15 \text{ cm} = 0.2 \text{ m} \times 0.15 \text{ m} = 0.03 \text{ m}^{2}$
$B = 2 \times 10^{-2} \text{ T}$
Frequency $f = 60 \text{ Hz}$
Angular velocity $\omega = 2 \pi f = 2 \pi \times 60 = 120 \pi \text{ rad/s}$
Substituting these values into the formula:
$\varepsilon_{0} = 250 \times 0.03 \times (2 \times 10^{-2}) \times 120 \pi$
Calculating the product:
$\varepsilon_{0} = 250 \times 0.03 \times 0.02 \times 120 \pi$
$\varepsilon_{0} = 7.5 \times 0.02 \times 120 \pi$
$\varepsilon_{0} = 0.15 \times 120 \pi$
$\varepsilon_{0} = 18 \pi \text{ volt}$
Answer: $18 \pi$ volt
Hence, the correct answer is option: 'B'.