Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To find the induced current, we must find the induced emf first.
The induced emf for a wire that moves in a magnetic field is given by
$$\varepsilon=vlB$$
Plug the known, and refer to Figure 33.26.
$$\varepsilon=(10)(0.2)(0.1)=\color{red}{\bf 0.2}\;\rm V$$
The induced current is given by
$$I=\dfrac{\varepsilon}{R}$$
$$I=\dfrac{0.2}{1}=\color{red}{\bf 0.2}\;\rm A$$
$$\color{blue}{\bf [b]}$$
The wire is moving at a constant speed which means that the pulling force equals the magnetic force exerted on this wire.
$$F_{\rm pull}=F_B$$
We know that the magnetic force exerted on a moving wire is given by
$$F_{\rm pull} =BIL=(0.1)(0.2)(0.2)$$
$$F_{\rm pull} =\color{red}{\bf 4}\;\rm mN$$
$$\color{blue}{\bf [c]}$$
We know that the power dissipated in the carbon resistor is given by
$$P=I^2R$$
where $P=Q/t$ where $Q$ is the amount of energy dissipated by the current and $t$ is the time interval. So
$$I^2R=\dfrac{Q}{t}$$
Recalling that $Q=mc\Delta T$,
$$I^2R=\dfrac{mc\Delta T}{t}$$
Hence,
$$\Delta T=\dfrac{I^2Rt}{mc}$$
Plug the known;
$$\Delta T=\dfrac{(0.2)^2(1.0)(10)}{(50\times 10^-6)(710)}$$
$$\Delta T=\color{red}{\bf 11.3}\;\rm K$$
