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1. Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} \oint \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} \oint \mathbf{E \cdot ds} = \int_{cond} \mathbf{E \cdot ds} + \int_{v-mtr} \mathbf{E \cdot ds} = 0 +\int_{v-mtr} \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

1. Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} emf = -\oint \mathbf{E \cdot ds}=\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} -\oint \mathbf{E \cdot ds} = -\int_{cond} \mathbf{E \cdot ds} - \int_{v-mtr} \mathbf{E \cdot ds} = 0 -\int_{v-mtr} \mathbf{E \cdot ds}=\frac{d\Phi}{dt} \end{equation}

1. Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} \oint \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} \oint \mathbf{E \cdot ds} = \oint_{cond} \mathbf{E \cdot ds} + \oint_{v-mtr} \mathbf{E \cdot ds} = 0 +\oint_{v-mtr} \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

1. Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} \oint \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} \oint \mathbf{E \cdot ds} = \int_{cond} \mathbf{E \cdot ds} + \int_{v-mtr} \mathbf{E \cdot ds} = 0 +\int_{v-mtr} \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

Yes. The electric field within the conductor is 0, but the voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

1. Yes there is a deflection. 2) The voltmeter will measure the time rate of change of the magnetic flux enclosed by the conductor + voltmeter circuit (Faraday's law).

Elaboration: Faraday's law says that the line integral of the electric field around a loop (loop emf), is equal to the time rate of change of the magnetic flux enclosed by the loop \begin{equation} \oint \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}

For this particular loop, consisting of a conductor + a voltmeter:

1. the contribution to the line integral from the part of the loop within the conductor is 0, by definition of a conductor. (The charges within the conductor distribute themselves so as to null its electric field.)
2. therefore the remainder of the line integral (the voltmeter portion of the circuit) must be the full loop emf:

\begin{equation} \oint \mathbf{E \cdot ds} = \oint_{cond} \mathbf{E \cdot ds} + \oint_{v-mtr} \mathbf{E \cdot ds} = 0 +\oint_{v-mtr} \mathbf{E \cdot ds}=-\frac{d\Phi}{dt} \end{equation}