A constant magnetic field of 1 T is applied in the x0 region. A metallic circular ring of radius 1

Q: Why can't we just use Faraday's Law by differentiating the area?

You absolutely can! However, calculating the derivative of the area of a circular segment with respect to time involves complex trigonometry. The motional EMF approach ($B l_{eff} v$) is a mathematically equivalent and much faster shortcut.

Q: Why is the effective length equal to the diameter?

At $t = 1 \text{ s}$, the center of the ring is at $x = 0$. The points of the ring that are currently intersecting the magnetic field boundary ($x = 0$) are the topmost and bottommost points. The straight distance between them is the diameter, $2R$.

Q: What if the magnetic field was not uniform?

If the magnetic field varied with position, we would have to integrate the motional EMF ($d\varepsilon = B(y) \cdot v \cdot dy$) along the effective length. The simple $B l_{eff} v$ formula only works for uniform fields.

Q: Does the resistance of the metallic ring affect the induced EMF?

No. The induced EMF depends only on the magnetic field, velocity, and geometry ($B, l, v$). However, if you were asked to find the induced current, then you would need the ring's resistance ($I = \varepsilon / R$).

Q: What happens to the EMF when the ring is completely inside the magnetic field?

Once the ring is fully inside, the magnetic flux through it becomes constant. Since the rate of change of flux is zero, the induced EMF instantly becomes zero.

Question

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

Options:

(A) $1	ext{ V}$

(B) $2\pi	ext{ V}$

(C) $2	ext{ V}$

(D) $0$

Sigma Percentile · Accepted Answer

The problem asks us to find the induced EMF in a metallic ring moving into a magnetic field. This is a classic application of **Faraday's Law** and the concept of **Motional EMF**.

### Visualizing the Journey

Imagine you are observing this metallic ring from above. At $t = 0 	ext{ s}$, the center of the ring is at $x = -1 	ext{ m}$. The ring has a radius of $1 	ext{ m}$, which means its rightmost edge is just touching the boundary of the magnetic field at $x = 0$.

The ring is moving with a constant velocity of $1 	ext{ m/s}$. Where will it be after exactly $1 	ext{ s}$?

Using basic kinematics, the distance traveled is:

$$d = v 	imes t = 1 	imes 1 = 1 	ext{ m}$$

So, the new center of the ring is at $x = -1 + 1 = 0 	ext{ m}$. At this exact moment, the ring is halfway inside the magnetic field.

### The Master Equation: Motional EMF

As the ring moves, the area of the ring inside the magnetic field is continuously increasing. This changing area leads to a changing magnetic flux, which in turn induces an EMF.

While we could use Faraday's Law ($\varepsilon = -\frac{d\Phi}{dt}$) by calculating the rate of change of the circular segment's area, there is a much more elegant tool: **Motional EMF**.

The formula for motional EMF is:

$$\varepsilon = B l_{eff} v$$

Here is the catch: $l_{eff}$ is not the actual curved length of the wire inside the field. It is the **effective length**—the straight-line distance between the two points of the conductor that are actively cutting the magnetic field boundary.

### Final Calculation

At $t = 1 	ext{ s}$, the points of the ring crossing the $x = 0$ boundary are exactly at the top and bottom of the ring. The straight-line distance between these two points is simply the diameter of the ring.

$$l_{eff} = 2R = 2(1) = 2 	ext{ m}$$

Now, we just substitute our known values into the motional EMF equation. We know the magnetic field $B = 1 	ext{ T}$, the effective length $l_{eff} = 2 	ext{ m}$, and the velocity $v = 1 	ext{ m/s}$.

$$\varepsilon = 1 	imes 2 	imes 1 = 2 	ext{ V}$$

The induced EMF in the ring at that instant is exactly $2 	ext{ V}$.

Always remember, if the ring were completely inside the uniform magnetic field, the effective length cutting the boundary would be zero, and the induced EMF would instantly drop to zero!

Select Answer:

Visualized Solution

Initial Position \& Velocity

Position at $t = 1 s$

Magnetic Flux Change

Motional EMF Formula

Effective Length

Final Calculation

Conceptual Takeaway

The Sigma Insight: Motional EMF

Visualizing the Journey

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Initial Position \& Velocity

Position at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=1 s

Magnetic Flux Change

Motional EMF Formula

Effective Length

Final Calculation

Conceptual Takeaway

The Sigma Insight: Motional EMF

Visualizing the Journey

The Master Equation

Final Calculation

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Position at $t = 1 s$