A straight wire of length L carries a steady current I along the...

The magnetic force on a straight current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.

Since the current flows along the negative y-axis, the length vector is $\vec{L} = -L\hat{j}$.

Let the magnetic field be $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$.

Substituting these into the force equation:

$\vec{F} = I(-L\hat{j}) \times (B_x\hat{i} + B_y\hat{j} + B_z\hat{k})$

Using the properties of cross products ($\hat{j} \times \hat{i} = -\hat{k}$, $\hat{j} \times \hat{j} = 0$, $\hat{j} \times \hat{k} = \hat{i}$):

$\vec{F} = -IL(B_x(\hat{j} \times \hat{i}) + B_y(\hat{j} \times \hat{j}) + B_z(\hat{j} \times \hat{k}))$

$\vec{F} = -IL(-B_x\hat{k} + B_z\hat{i}) = IL(-B_z\hat{i} + B_x\hat{k})$

We are given that the force is $\vec{F} = IL(3\hat{i} + 4\hat{k})$.

Comparing the components of the calculated force with the given force:

$-B_z = 3 \Rightarrow B_z = -3$

$B_x = 4$

The y-component $B_y$ can be any real number since the force does not depend on the magnetic field component parallel to the wire.

Therefore, the magnetic field must be of the form $\vec{B} = 4\hat{i} + B_y\hat{j} - 3\hat{k}$.

Among the given options, only $(4\hat{i} + 6\hat{j} - 3\hat{k})$ satisfies these conditions.

Answer: $(4\hat{i} + 6\hat{j} - 3\hat{k})$