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$$^{-1}$$ 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 = 1 s? (Assume the velocity of the ring does not change.)

We need to determine the induced electromotive force (e.m.f.) in a metallic circular ring moving along the $$x$$-axis through a boundary where a magnetic field begins.

1. Analyze the Motion of the Ring

Let's map out the position of the ring based on the given parameters:

• Radius of the ring ($$R$$): $$1\text{ m}$$
• Constant velocity ($$v$$): $$1\text{ m s}^{-1}$$ along the positive $$x$$-axis.
• Initial Position ($$t = 0\text{ s}$$): The center $$O$$ of the ring is at $$x = -1\text{ m}$$.

We want to find the state of the ring at time $$t = 1\text{ s}$$. Using the equation of motion:

$$\text{Displacement} = v \times t = 1\text{ m s}^{-1} \times 1\text{ s} = 1\text{ m}$$

Therefore, at $$t = 1\text{ s}$$, the center $$O$$ of the ring has shifted from $$x = -1\text{ m}$$ to:

$$x = -1\text{ m} + 1\text{ m} = 0\text{ m}$$

2. Determine the Position Relative to the Magnetic Field

The magnetic field ($$B = 1\text{ T}$$) is applied only in the region where $$x > 0$$. At $$t = 1\text{ s}$$, since the center of the ring is exactly at $$x = 0$$:

• The right half of the circular ring (from $$x = 0$$ to $$x = +1\text{ m}$$) is inside the magnetic field region.
• The left half of the circular ring (from $$x = -1\text{ m}$$ to $$x = 0$$) is outside the magnetic field region.

3. Calculate the Induced E.M.F. (Motional E.M.F.)

As the ring crosses the boundary, the change in magnetic flux induces an e.m.f. This can be effectively evaluated using the motional e.m.f. formula for a conductor cutting through a magnetic field:

$$e = B \cdot l_{\text{eff}} \cdot v$$

Where:

• $$B = 1\text{ T}$$ (Magnetic field intensity)
• $$v = 1\text{ m s}^{-1}$$ (Velocity of the conductor)
• $$l_{\text{eff}}$$ is the effective length of the vertical segment of the conductor that cuts the magnetic boundary lines perpendicularly.

The boundary lies along the vertical $$y$$-axis at $$x = 0$$. The segment of the ring cutting through this boundary spans from the bottom-most point to the top-most point along the line $$x = 0$$. This vertical distance is equal to the full diameter of the circular ring:

$$l_{\text{eff}} = 2R = 2 \times 1\text{ m} = 2\text{ m}$$

Substituting these values into the motional e.m.f. formula:

$$e = 1\text{ T} \times 2\text{ m} \times 1\text{ m s}^{-1} = 2\text{ V}$$

Final Answer: 2 V (Option B)