Step-by-step solution:

1. Identify the relevant physical principles. The problem involves a changing magnetic field creating an induced electric field. This phenomenon is described by Faraday's Law of Induction. The law relates the induced electromotive force (EMF) to the rate of change of magnetic flux. The EMF is also related to the line integral of the induced electric field around a closed loop.

2. Faraday's Law of Induction. Faraday's Law states that the induced EMF, ε, in a closed loop is equal to the negative time rate of change of the magnetic flux, $Φ_B$, through the loop. $ε = -{{dΦ_B} \over {dt}}$

3. Calculate the Magnetic Flux ($Φ_B$). The charge moves in a circular orbit of radius R in the xy-plane. The area of this orbit (loop) is A = πR². The magnetic field B(t) is uniform and directed along the positive z-axis, which is perpendicular to the plane of the orbit. Therefore, the magnetic flux through the loop is: $Φ_B = \vec{B} \cdot \vec{A} = B(t) A \cos(0°) = B(t) πR²$

4. Calculate the Rate of Change of Magnetic Flux. We can now substitute the expression for $Φ_B$ into Faraday's Law. The problem states that the radius R of the orbit remains constant. $ε = -{{d} \over {dt}}(B(t) πR²) = -πR² {{dB} \over {dt}}$

5. Determine the Rate of Change of the Magnetic Field (dB/dt). The magnetic field increases at a constant rate from 0 to a final magnitude B in one second. Thus, the rate of change dB/dt is constant: ${{dB} \over {dt}} = {{ΔB} \over {Δt}} = {{B_{final} - B_{initial}} \over {t_{final} - t_{initial}}} = {{B - 0} \over {1 - 0}} = B$ The units are Tesla/second.

6. Calculate the Magnitude of the Induced EMF. Substituting the value of dB/dt into the equation for EMF, we find the magnitude of the induced EMF: $|ε| = |-πR² B| = πR² B$

7. Relate EMF to the Induced Electric Field (E). The problem defines the induced EMF as the work done by the induced electric field (E) in moving a unit positive charge around the closed loop. This is mathematically expressed as the line integral of the electric field around the loop: $ε = \oint \vec{E} \cdot d\vec{l}$ Due to the cylindrical symmetry of the setup (uniform magnetic field along the axis of a circular loop), the induced electric field lines must be concentric circles in the xy-plane. The magnitude of the electric field, E, will be constant at all points on the circular orbit of radius R. The vector $\vec{E}$ is tangential to the orbit, and so is the line element $d\vec{l}$. Thus, $\vec{E} \cdot d\vec{l} = E dl$. The integral becomes: $|ε| = \oint E dl = E \oint dl = E \times (\text{circumference of the orbit})$ $|ε| = E (2πR)$

8. Solve for the Magnitude of the Induced Electric Field (E). We now have two expressions for the magnitude of the induced EMF. By equating them, we can solve for E: $E (2πR) = πR² B$ $E = {{πR² B} \over {2πR}}$ $E = {{BR} \over {2}}$ This is the magnitude of the induced electric field at any instant during the one-second interval, as dB/dt is constant.

9. Compare with Options. The calculated value E = BR/2 matches option B.

The extra information about the charge Q, angular velocity ω, and the proportionality between magnetic dipole moment and angular momentum is not needed to solve for the induced electric field and can be disregarded for this specific question.