A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is direct

Q: Why do we take the area as $\pi a^2$ and not $\pi r^2$ when calculating flux for the outside point?

Because the magnetic field only exists up to radius $a$. The region between $a$ and $r$ has absolutely no magnetic field, so it contributes zero to the total magnetic flux passing through the loop of radius $r$.

Q: Why is the induced electric field tangential to the circular loop?

Due to the cylindrical symmetry of the changing magnetic field, the induced electric field lines form closed concentric circles. The field must be tangential to these circles everywhere to satisfy the symmetry conditions.

Q: Is this induced electric field conservative like electrostatic fields?

No, induced electric fields produced by changing magnetic fields are non-conservative. The line integral around a closed loop is not zero (it equals the rate of change of flux), which is exactly why they can drive a continuous current in a closed circuit.

Q: What happens to the electric field exactly at the boundary $r = a$?

The electric field is perfectly continuous at the boundary. If you plug $r = a$ into both the inside formula ($E \propto r$) and the outside formula ($E \propto 1/r$), you will get the exact same maximum value: $E = \frac{a}{2} \left| \frac{dB}{dt} \right|$.

Q: Does the direction of the magnetic field matter for the magnitude of $E$?

The direction of the magnetic field (into or out of the page) determines the *direction* of the induced electric field (clockwise or counter-clockwise, via Lenz's Law), but the *magnitude* depends only on the rate of change $\left| \frac{dB}{dt} \right|$.

Question

A uniform but time-varying magnetic field $B(t)$ exists in a circular region of radius $a$ and is directed into the plane of the paper as shown. The magnitude of the induced electric field at point $P$ at a distance $r$ from the centre of the circular region

Options:

(A) is zero

(B) decreases as $1/r$

(C) increases as $r$

(D) decreases as $1/r^2$

Sigma Percentile · Accepted Answer

### The Magic of Changing Magnetic Fields

Imagine a cylindrical region in space where a magnetic field is constantly changing with time. According to Faraday's Law of Induction, this changing magnetic field doesn't just sit there—it actively creates an electric field around it. Our mission in this problem is to find the magnitude of this induced electric field at a point $P$, which is located at a distance $r$ outside the magnetic region (where $r > a$).

### The Master Tool: Faraday's Law

To solve this, we need to bring out our heavy artillery: Faraday's Law of Induction in its integral form. It states that the line integral of the electric field around a closed loop is equal to the negative rate of change of magnetic flux through that loop:

$$\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$$

Let's draw an imaginary circular loop (an Amperian loop) of radius $r$ passing right through our point $P$. Because of the perfect cylindrical symmetry of the setup, the induced electric field $\vec{E}$ will be tangential to this loop at every single point. Furthermore, its magnitude will be constant everywhere on the loop.

### Evaluating the Left-Hand Side

Now, let's evaluate the left side of Faraday's law. Since the electric field $\vec{E}$ is parallel to the path element $d\vec{l}$, the dot product is simply $E \, dl$. And because $E$ is constant, we can pull it out of the integral. The integral of $dl$ over the whole circle is just its circumference, $2\pi r$.

$$\oint \vec{E} \cdot d\vec{l} = E \oint dl = E(2\pi r)$$

### The Catch: Evaluating the Magnetic Flux

Next, let's look at the right side: the magnetic flux $\Phi_B$. Here is a crucial catch! Even though our imaginary loop has a radius $r$, the magnetic field only exists up to a radius $a$. The space between $a$ and $r$ is completely empty of magnetic field lines.

So, the area that actually contains magnetic flux is just $\pi a^2$, not $\pi r^2$. The rate of change of flux is then $\pi a^2$ times the rate of change of the magnetic field.

$$\left| \frac{d\Phi_B}{dt} ight| = \pi a^2 \left| \frac{dB}{dt} ight|$$

### The Final Calculation

Equating both sides, we get:

$$E(2\pi r) = \pi a^2 \left| \frac{dB}{dt} ight|$$

The $\pi$ cancels out beautifully, and we can solve for $E$:

$$E = \frac{a^2}{2r} \left| \frac{dB}{dt} ight|$$

From this expression, it is crystal clear that the induced electric field $E$ is inversely proportional to the distance $r$.

**$	herefore E \propto \frac{1}{r}$**

So, as we move further away from the magnetic region, the electric field decreases as $1/r$.

### The Way Forward: What if we were inside?

What if our point $P$ was *inside* the magnetic region ($r < a$)? In that case, the flux would be enclosed by the entire area of our smaller loop, $\pi r^2$.

$$E(2\pi r) = \pi r^2 \left| \frac{dB}{dt} ight| \implies E = \frac{r}{2} \left| \frac{dB}{dt} ight|$$

This gives an electric field that is directly proportional to $r$ ($E \propto r$). Inside the region, the field increases linearly from the center, and outside, it drops off as $1/r$. This is a classic, high-yield profile that you should permanently etch into your memory!

Select Answer:

Visualized Solution

$Visual Anchor: The Setup$

$Logic Bridge: Faraday’s Law$

$Raw Setup: Amperian Loop$

$Atomic Compute: Line Integral$

$Atomic Compute: Magnetic Flux$

$Final Answer: Outside Field$

$The Way Forward: Inside Field$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Magic of Changing Magnetic Fields

The Master Tool

Evaluating the Left-Hand Side

The Catch

The Final Calculation

The Way Forward

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