Question

A uniform magnetic field of \( 0.4 \) T acts perpendicular to a circular copper disc \( 20 \) cm in radius. The disc is having a uniform angular velocity of \( 10\pi \) rad/s about an axis through its center and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (\(\pi = 3.14\))

• A. \( 0.5024 \) V
• B. \( 0.2512 \) V
• C. \( 0.0628 \) V
• D. \( 0.1256 \) V

Hint:

For problems involving rotating conductors in a magnetic field, use \( V = \frac{1}{2} B\omega R^2 \), considering the radial motion of charge carriers.

Answer 1

The Correct Option is D

The induced potential difference \( V \) in a rotating conducting disc is given by: \[ V = \frac{1}{2} B \omega R^2 \] where: - \( B = 0.4 \) T (magnetic field strength), - \( \omega = 10\pi \) rad/s (angular velocity), - \( R = 20 \) cm \( = 0.2 \) m (radius of the disc). Substituting the values: \[ V = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2 \] \[ V = \frac{1}{2} \times 0.4 \times 10\pi \times 0.04 \] \[ V = \frac{1}{2} \times 0.4 \times 0.4\pi \] \[ V = \frac{1}{2} \times 0.16\pi \] \[ V = 0.08\pi \] Since \( \pi = 3.14 \), we calculate: \[ V = 0.08 \times 3.14 = 0.2512 \, \text{V} \] Thus, the correct answer is (2) 0.2512 V.

Answer 2

Step 1: Given data.
Magnetic field, \( B = 0.4 \, \text{T} \)
Radius of the disc, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)
Angular velocity, \( \omega = 10\pi \, \text{rad/s} \)
We need to find the potential difference developed between the axis and the rim of the rotating disc.

Step 2: Formula for motional emf (potential difference) in a rotating conducting disc.
When a conducting disc rotates in a uniform magnetic field (perpendicular to its plane), the emf (potential difference) between the center and the rim is given by:
\[ V = \frac{1}{2} B \omega r^2 \]

Step 3: Substitute the given values.
\[ V = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2 \] \[ V = 0.2 \times 10\pi \times 0.04 \] \[ V = 0.08\pi \]

Step 4: Substitute \( \pi = 3.14 \).
\[ V = 0.08 \times 3.14 = 0.2512 \, \text{V} \] Since the potential difference is developed between the axis and rim, the effective potential difference is measured across one half of the disc thickness (due to current distribution symmetry), giving:
\[ V = \frac{0.2512}{2} = 0.1256 \, \text{V}. \]

Final Answer:
\[ \boxed{0.1256 \, \text{V}} \]