A conductor is placed along z-axis carrying current in z direction in uniform magnetic field directed along y-axis. The magnetic force acting on the conductor is directed along:

The correct option is negative x-axis.

To understand why this is the correct direction, we can use the formula for the magnetic force acting on a current-carrying conductor placed in a uniform magnetic field. The magnetic force $\overrightarrow{F}$ is given by the expression:
$\overrightarrow{F}=I(\overrightarrow{L}\times \overrightarrow{B})$
where:
• $I$ is the current flowing through the conductor,
• $\overrightarrow{L}$ is the length vector of the conductor pointing in the direction of the current, and
• $\overrightarrow{B}$ is the magnetic field vector.

Let us write down the unit vectors along the coordinate axes:
• The unit vector along the positive x-axis is $\hat{i}$.
• The unit vector along the positive y-axis is $\hat{j}$.
• The unit vector along the positive z-axis is $\hat{k}$.

According to the given problem statement:
1. The conductor is carrying current along the z-direction. Therefore, the direction of the length vector $\overrightarrow{L}$ is along the positive z-axis:
$\overrightarrow{L}=L\hat{k}$
2. The uniform magnetic field is directed along the y-axis (positive y-axis assumed unless specified otherwise):
$\overrightarrow{B}=B\hat{j}$

Substitute these values into the magnetic force formula:
$\overrightarrow{F}=I(L\hat{k}\times B\hat{j})$
$\overrightarrow{F}=ILB(\hat{k}\times \hat{j})$

By using the vector cross product rules for unit vectors, we know that:
$\hat{k}\times \hat{j}=-\hat{i}$

Therefore, substituting this back into the force equation gives:
$\overrightarrow{F}=-ILB\hat{i}$

Since the unit vector is $-\hat{i}$, the direction of the magnetic force is along the negative x-axis (or alternatively, using Right-Hand Rule: pointing fingers in the +z direction and curling them toward +y points the thumb in the -x direction).