Magnitude of Induced Electric Field in Changing Magnetic Field

To find the magnitude of the electric field induced in the circular orbit during the period in which the magnetic field is changing, we use Faraday's Law of Electromagnetic Induction.

1. Rate of Change of Magnetic Field

The magnetic field increases at a uniform rate from $0$ to $B$ over a duration of $\Delta t = 1\text{ second}$. Therefore, the rate of change of the magnetic field with respect to time, $\frac{dB}{dt}$, is: $\frac{dB}{dt} = \frac{B - 0}{1} = B$

2. Magnetic Flux Through the Orbit

The circular orbit has a constant radius $R$. The area enclosed by the circular path is: $A = \pi R^2$

Since the magnetic field is uniform and perpendicular to the $x-y$ plane (parallel to the $z$-axis), the magnetic flux $\Phi_B$ through the loop is: $\Phi_B = B \cdot A = B (\pi R^2)$

The rate of change of magnetic flux is given by: $\frac{d\Phi_B}{dt} = A \frac{dB}{dt} = \pi R^2 B$

3. Induced Electromotive Force (emf) and Induced Electric Field

According to Faraday's Law, the magnitude of the induced electromotive force ($\mathcal{E}$) around the circular orbit is equal to the rate of change of magnetic flux: $\mathcal{E} = \left| -\frac{d\Phi_B}{dt} \right| = \pi R^2 B$

The induced emf is also defined as the line integral of the induced electric field $\vec{E}$ along the circular loop: $\mathcal{E} = \oint \vec{E} \cdot d\vec{l}$

Due to cylindrical symmetry, the magnitude of the induced electric field $E$ is constant at every point along the circular path of radius $R$, and its direction is tangential to the path. Thus: $\mathcal{E} = E \oint dl = E (2\pi R)$

4. Equating both Expressions for Induced emf

$E (2\pi R) = \pi R^2 B$

Solving for $E$: $E = \frac{\pi R^2 B}{2\pi R} = \frac{BR}{2}$

Thus, the magnitude of the induced electric field at any instant during this time interval is $\frac{BR}{2}$.

Correct Option: B