A horizontal ring of radius `r` spins about its axis with an angular velocity `omega` in a uniform vertical magnetic field of magnitude `B`. The emf induced in the ring is

Options

A. zero
B. `pir^(2) omegaB`
C. `1/2 Br^(2) omega`
D. `Br^(2) omega`

Answer

To solve the problem of finding the induced emf in a horizontal ring spinning in a uniform vertical magnetic field, we can follow these steps:

### Step 1: Understand the Setup
We have a horizontal ring of radius \( r \) that is spinning about its axis with an angular velocity \( \omega \). The ring is placed in a uniform vertical magnetic field of magnitude \( B \).

### Step 2: Analyze the Magnetic Flux
The magnetic flux \( \Phi \) through the ring is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where:
- \( B \) is the magnetic field strength,
- \( A \) is the area of the ring,
- \( \theta \) is the angle between the magnetic field direction and the normal to the surface of the ring.

Since the ring is horizontal and the magnetic field is vertical, the angle \( \theta \) is \( 90^\circ \). Thus, \( \cos(90^\circ) = 0 \).

### Step 3: Calculate the Area of the Ring
The area \( A \) of the ring can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]

### Step 4: Determine the Magnetic Flux
Substituting the area into the flux formula:
\[ \Phi = B \cdot \pi r^2 \cdot \cos(90^\circ) = B \cdot \pi r^2 \cdot 0 = 0 \]
So, the magnetic flux through the ring is zero.

### Step 5: Apply Faraday's Law of Electromagnetic Induction
Faraday's law states that the induced emf \( \mathcal{E} \) in a closed loop is equal to the negative rate of change of magnetic flux through the loop:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Since the magnetic flux \( \Phi \) is zero and does not change with time (as both \( B \) and \( A \) are constant and \( \theta = 90^\circ \)), we have:
\[ \frac{d\Phi}{dt} = 0 \]
Thus, the induced emf is:
\[ \mathcal{E} = -0 = 0 \]

### Conclusion
The induced emf in the ring is zero.