8.
A $0.75 \mathrm{~m}$ conducting rod is moved at $8.0 \mathrm{~m} / \mathrm{s}$ across a $0.25 \mathrm{~T}$ magnetic field along metal rails. The electrical resistance of the system is $5.0 \Omega$.
What are the magnitude and direction of the current through point $\mathrm{X}$ ?
A.
B.
C.
D.
\begin{tabular}{|c|c|}
\hline MAGNITUDE OF CURRENT & DIRECTION OF CURRENT THROUGH X \\
\hline $0.16 \mathrm{~A}$ & Left \\
\hline $0.16 \mathrm{~A}$ & Right \\
\hline $0.30 \mathrm{~A}$ & Left \\
\hline $0.30 \mathrm{~A}$ & Right \\
\hline
\end{tabular}

Step 1 ：Given values: B = 0.25 T, L = 0.75 m, v = 8.0 m/s, R = 5.0 Ω

Step 2 ：Calculate the induced EMF using the formula EMF = B * L * v

Step 3 ：EMF = 0.25 T * 0.75 m * 8.0 m/s = \(1.5 V\)

Step 4 ：Calculate the current using Ohm's law: I = EMF / R

Step 5 ：I = \(\frac{1.5 V}{5.0 Ω}\) = \(0.3 A\)

Step 6 ：Use the right-hand rule to find the direction of the current

Step 7 ：Since the magnetic field is going into the plane of the paper, and the rod is moving to the right, the current will be going to the left

Step 8 ：\(\boxed{\text{Final Answer: The magnitude of the current is 0.3 A, and the direction of the current through point X is to the left.}}\)