A ring made of insulating material is rolling without slipping on a horizontal surface with velocity of centre of mass `V_(0)`. A conducting wire of length 2R (R = radius of ring) is fixed between two points of the circumference. At an instant, the wire is in vertical position as shown in figure. A uniform magnetic field B exists perpendicular to the plane of the ring. The magnitude to emf induced between the ends of wire is

Options

A. `2BV_(0)R`
B. `BV_(0)R`
C. `3BV_(0)R`
D. `4BV_(0)R`

Answer

To solve the problem, we need to determine the induced electromotive force (emf) in the conducting wire fixed on the circumference of the rolling ring. We will use the formula for induced emf in a conductor moving through a magnetic field.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Let the radius of the ring be \( R \).
- The length of the conducting wire is \( 2R \).
- The velocity of the center of mass of the ring is \( V_0 \).
- The magnetic field \( B \) is uniform and perpendicular to the plane of the ring.

2. **Understand the motion:**
- The ring is rolling without slipping. This means that the point of contact with the ground is momentarily at rest while the center of mass moves with velocity \( V_0 \).

3. **Use the formula for induced emf:**
- The induced emf (\( \mathcal{E} \)) in a wire moving through a magnetic field is given by:
\[ \mathcal{E} = V \cdot B \cdot L \]
- Where:
- \( V \) is the velocity of the wire (which is the same as the velocity of the center of mass, \( V_0 \)).
- \( B \) is the magnetic field strength.
- \( L \) is the length of the wire.

4. **Substitute the known values:**
- Here, \( L = 2R \) (the length of the conducting wire).
- Therefore, substituting the values into the formula:
\[ \mathcal{E} = V_0 \cdot B \cdot (2R) \]

5. **Simplify the expression:**
- This simplifies to:
\[ \mathcal{E} = 2 V_0 B R \]

6. **Conclusion:**
- The magnitude of the induced emf between the ends of the wire is:
\[ \mathcal{E} = 2 V_0 B R \]

### Final Answer:
The magnitude of the emf induced between the ends of the wire is \( 2 V_0 B R \).

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