Key idea: motional emf in the moving rod

The rod of length $L$ moves left with speed $v$ in a uniform magnetic field $\vec B$ directed into the page.

Charges in the rod experience magnetic force $\vec F = q(\vec v \times \vec B).$

Here:

• $\vec v$ is toward the left
• $\vec B$ is into the page

Therefore, $\vec v \times \vec B$ points downward along the rod.

So positive charges are pushed downward, meaning:

• lower end of the rod is at higher potential
• upper end of the rod is at lower potential

Equivalent source polarity across the rails

Since the moving bar connects the two rails, it acts like a source of motional emf with:

• bottom rail at higher potential
• top rail at lower potential

Hence current in each external branch connected across the rails will flow from bottom rail to top rail through the resistor.

Direction of current through each resistor

There are two resistor branches across the rails: $R_1$ and $R_2$.

Since conventional current goes from higher to lower potential externally, in each resistor branch current goes: $\text{bottom rail} \to \text{resistor} \to \text{top rail}.$

Now determine loop sense:

• For the left branch (with $R_1$), current goes up through the resistor, then along the top rail to the moving rod, down the rod, and back along the bottom rail. This circulation is clockwise.
• For the right branch (with $R_2$), current goes up through the resistor, then along the top rail, down the rod, and back along the bottom rail. This circulation is anticlockwise.

Check with Lenz's law

As the rod moves left, the area of the left loop decreases, so flux into the page through the left loop decreases. Induced current must produce field into the page, which requires a clockwise current in the left loop.

For the right loop, area increases, so flux into the page increases. Induced current must oppose this increase by producing field out of the page, which requires an anticlockwise current in the right loop.

This confirms the same result.

Conclusion

• $I_1$ is clockwise
• $I_2$ is anticlockwise

Therefore, the correct option is B.