Question

A constant magnetic field of 1 T is applied in the x \(>\) 0 region. A metallic circular ring of radius 1 m is moving with a constant velocity of 1 m/s along the x-axis. At t=0 s, the centre O of the ring is at x = -1 m. What will be the value of the induced emf in the ring at t=1s? (Assume the velocity of the ring does not change.) 

 

• A. 0 V
• B. 1 V
• C. 2 V
• D. 2\(\pi\) V

Hint:

When a symmetric loop (like a circle or square) enters a uniform magnetic field, the motional EMF can be quickly calculated by considering the effective length of the conductor cutting the flux lines, which is the length of the leading edge inside the field, perpendicular to the velocity. Here, it's the vertical diameter.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A circular conducting ring is entering a region with a uniform magnetic field. We need to find the induced electromotive force (emf) at a specific time.
Step 2: Key Formula or Approach:
The induced emf in a conductor moving in a magnetic field is called motional emf. For a straight conductor of length L moving with velocity v perpendicular to a magnetic field B, the emf is given by \( \varepsilon = BLv \). For a conducting loop entering a field, the changing magnetic flux also induces an emf given by Faraday's law, \( \varepsilon = -\frac{d\Phi_B}{dt} \). Both approaches lead to the same result. The motional emf approach is often simpler here.
Step 3: Detailed Explanation:
1. Position of the ring at t = 1 s:
- Initial position of the center (at t=0): \( x_0 = -1 \) m.
- Constant velocity: \( v = 1 \) m/s.
- Position at time t: \( x(t) = x_0 + vt \).
- At t = 1 s: \( x(1) = -1 + (1)(1) = 0 \) m.
So, at t = 1 s, the center of the ring is exactly at the boundary \(x=0\). This means the ring is halfway into the magnetic field region.
2. Calculating the Induced EMF:
As the ring enters the field, an emf is induced. We can think of the ring as being made of many small conducting segments. The emf is induced in the segment of the ring that is cutting the magnetic flux lines. This is the vertical chord of the ring that is currently at the boundary \(x=0\).
At \(t=1\)s, the diameter of the ring lies along the y-axis. The length of this conductor cutting the flux is the diameter of the ring.
- Effective length of the conductor, \( L = \text{Diameter} = 2 \times \text{Radius} = 2R \).
- Given Radius \( R = 1 \) m, so \( L = 2 \) m.
The velocity of this conductor is \( v = 1 \) m/s, and the magnetic field is \( B = 1 \) T. The velocity, length, and field are mutually perpendicular.
Using the motional emf formula:
\[ \varepsilon = B L v \] \[ \varepsilon = (1 \, \text{T}) \times (2 \, \text{m}) \times (1 \, \text{m/s}) \] \[ \varepsilon = 2 \, \text{V} \] The emfs in the top and bottom semicircles add up to produce this total emf across the loop's diameter.
Step 4: Final Answer:
The value of the induced emf in the ring at t = 1 s is 2 V.