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(A) Statement $(I)$: The magnetic force on the electron is $\vec{F}_m = q(\vec{v} 	imes \vec{B}) = 0$ because the velocity $\vec{v}$ is parallel to t

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Statements $I$, $II$ and $III$ are true.

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(A) Statement $(I)$: The magnetic force on the electron is $\vec{F}_m = q(\vec{v} 	imes \vec{B}) = 0$ because the velocity $\vec{v}$ is parallel to the magnetic field $\vec{B}$. The electric force is $\vec{F}_e = q\vec{E} = -e\vec{E}$. Since the electron is negatively charged, the electric force acts opposite to the direction of the electric field. As the electron moves in the direction of the field, the electric force acts against its motion, causing its velocity to decrease. Thus, Statement $(I)$ is true.Statement $(II)$: For two parallel wires carrying current $i$ in the same direction, the magnetic field produced by each wire at the midpoint is equal in magnitude but opposite in direction (using the right-hand rule). Therefore, the net magnetic field $\vec{B}_{	ext{net}} = \vec{B}_1 + \vec{B}_2 = 0$. Thus, Statement $(II)$ is true.Statement $(III)$: For a rectangular coil carrying a steady current in a uniform magnetic field, the forces on opposite sides are equal and opposite, resulting in a net force of zero. Thus, Statement $(III)$ is true.Therefore, all statements are true.

Column $I$	Column $II$	Column $III$
$(I)$ Electron with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(i)$ $\overrightarrow{E}=E_0 \hat{z}$	$(P)$ $\overrightarrow{B}=-B_0 \hat{x}$
$(II)$ Electron with $\overrightarrow{v}=\frac{E_0}{B_0} \hat{y}$	$(ii)$ $\overrightarrow{E}=-E_0 \hat{y}$	$(Q)$ $\overrightarrow{B}=B_0 \hat{x}$
$(III)$ Proton with $\overrightarrow{v}=0$	$(iii)$ $\overrightarrow{E}=-E_0 \hat{x}$	$(R)$ $\overrightarrow{B}=B_0 \hat{y}$
$(IV)$ Proton with $\overrightarrow{v}=2 \frac{E_0}{B_0} \hat{x}$	$(iv)$ $\overrightarrow{E}=E_0 \hat{x}$	$(S)$ $\overrightarrow{B}=B_0 \hat{z}$

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