Question

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s\(^{-1}\) in a uniform horizontal magnetic field of 3.0\( \times \)10\(^{-2}\) T. The maximum emf induced the coil will be __________ \( \times \)10\(^{-2}\) volt (rounded off to the nearest integer).

Hint:

The formula for maximum induced EMF in a generator, \( \epsilon_{max} = NBA\omega \), is fundamental. Ensure all quantities are in SI units before calculation (e.g., convert cm to m). The final step of matching the required units and rounding is also crucial.

Answer 1

Correct Answer: 60

Step 1: Understanding the Question:
This question asks for the maximum electromotive force (emf) induced in a coil rotating in a uniform magnetic field. This is the basic principle of an AC generator.
Step 2: Key Formula or Approach:
The emf induced in a coil with N turns, area A, rotating with angular velocity \(\omega\) in a uniform magnetic field B is given by \( \epsilon = NBA\omega \sin(\omega t) \).
The maximum emf (\(\epsilon_{max}\)) occurs when \( \sin(\omega t) = 1 \).
Thus, \( \epsilon_{max} = NBA\omega \).
Step 3: Detailed Explanation:
Given values:
Number of turns, N = 20
Radius of the coil, r = 8.0 cm = 0.08 m
Angular speed, \(\omega\) = 50 rad s\(^{-1}\)
Magnetic field, B = 3.0 \( \times \) 10\(^{-2}\) T
First, calculate the area of the coil, A:
\[ A = \pi r^2 = \pi (0.08 \text{ m})^2 = 0.0064\pi \text{ m}^2 \] Now, use the formula for the maximum induced emf:
\[ \epsilon_{max} = N B A \omega \] \[ \epsilon_{max} = 20 \times (3.0 \times 10^{-2} \text{ T}) \times (0.0064\pi \text{ m}^2) \times (50 \text{ rad s}^{-1}) \] Let's group the terms for easier calculation:
\[ \epsilon_{max} = (20 \times 50) \times (3.0 \times 10^{-2}) \times (0.0064\pi) \] \[ \epsilon_{max} = 1000 \times (3.0 \times 10^{-2}) \times (0.0064\pi) \] \[ \epsilon_{max} = 30 \times 0.0064\pi = 0.192\pi \text{ V} \] Using the value of \( \pi \approx 3.14159 \):
\[ \epsilon_{max} \approx 0.192 \times 3.14159 \approx 0.60318 \text{ V} \] The question asks for the answer in units of \( \times 10^{-2} \) volt.
\[ \epsilon_{max} = 0.60318 \text{ V} = 60.318 \times 10^{-2} \text{ V} \] Rounding off to the nearest integer, we get 60.
Step 4: Final Answer:
The maximum emf induced in the coil is 60 \( \times \) 10\(^{-2}\) volt.