We are given the following details in the problem:
- A uniform magnetic field of induction B is confined to a cylindrical region of radius R.
- The magnetic field is increasing at a constant rate \(\dfrac{dB}{dt}\).
- An electron of charge q is placed at the point P on the periphery of the field.
To solve the problem, we need to find the acceleration experienced by the electron.
### Step 1: Understanding the situation
- The magnetic field \(B\) is increasing with time.
- The point P lies at the periphery of the field, so the radius of the circular field is \(R\).
- Since the magnetic field is increasing, it will induce an electric field due to Faraday’s Law of Induction.
### Step 2: Apply Faraday’s Law of Induction
Faraday's Law states that a changing magnetic flux induces an electric field. The induced electromotive force (EMF) is given by:
\[
\mathcal{E} = -\dfrac{d\Phi_B}{dt}
\]
where \(\Phi_B\) is the magnetic flux through the area enclosed by the periphery of the cylindrical region. The magnetic flux is given by:
\[
\Phi_B = B \cdot A = B \cdot \pi R^2
\]
Since \(B\) is changing with time, we have:
\[
\dfrac{d\Phi_B}{dt} = \pi R^2 \dfrac{dB}{dt}
\]
So, the induced EMF is:
\[
\mathcal{E} = - \pi R^2 \dfrac{dB}{dt}
\]
### Step 3: Induced electric field
The electric field \(\vec{E}\) induced in the region due to this changing magnetic field is related to the EMF. The EMF in a circular loop of radius \(R\) can be written as:
\[
\mathcal{E} = E \cdot 2\pi R
\]
where \(E\) is the magnitude of the induced electric field. Equating the two expressions for the EMF:
\[
E \cdot 2\pi R = \pi R^2 \dfrac{dB}{dt}
\]
Simplifying:
\[
E = \dfrac{R}{2} \dfrac{dB}{dt}
\]
### Step 4: Force on the electron
The force on the electron due to the induced electric field is:
\[
F = qE
\]
Substituting the value of \(E\):
\[
F = q \cdot \dfrac{R}{2} \dfrac{dB}{dt}
\]
### Step 5: Acceleration of the electron
The acceleration of the electron can be found using Newton's second law:
\[
a = \dfrac{F}{m}
\]
where \(m\) is the mass of the electron. Substituting the expression for \(F\):
\[
a = \dfrac{q \cdot \dfrac{R}{2} \dfrac{dB}{dt}}{m}
\]
This simplifies to:
\[
a = \dfrac{q R}{2m} \dfrac{dB}{dt}
\]
### Step 6: Direction of acceleration
The direction of the induced electric field follows the right-hand rule. Since the magnetic field is increasing, the induced electric field will point in a direction tangential to the circular region. If the electron is at the periphery of the field, the force (and thus the acceleration) will be directed either toward the left or right depending on the sign of the charge.
Since the electron has a negative charge (\(q = -e\)), the force will be directed opposite to the direction of the electric field. Therefore, the electron will experience an acceleration toward the left.
### Final Answer:
The acceleration experienced by the electron is:
\[
\boxed{\dfrac{1}{2} \dfrac{e R}{m} \dfrac{dB}{dt} \text{ toward left}}
\]
Thus, the correct option is **(A)**.
