Question

a metal cube with sides of length ( a = 1 \text{ cm} ) is moving at velocity ( vec{v}_0 = 1 \text{ m/s} hat{j} ) across a uniform magnetic field ( vec{b}_0 = 5 \text{ t} hat{k} ). the cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vectors of two faces are parallel to the direction of motion) (figure 1).

Explanation:

According to the Lorentz force formula \( \vec{F} = q(\vec{v}\times\vec{B}) \), for a positive charge in the metal cube, the velocity \( \vec{v}=v_0\vec{j} \) (assuming the cube is moving along the \( y - \)axis as per the standard coordinate system in such problems, since \( \vec{v}_0 = 1m/s\vec{j} \) is a common setup when the cube's motion is described with one - axis velocity in the given context of magnetic field \( \vec{B}_0=B_0\vec{k} \)). The cross - product \( \vec{v}\times\vec{B}=(v_0\vec{j})\times(B_0\vec{k})=v_0B_0\vec{i} \). The induced electric field \( \vec{E} \) is related to the force on a charge \( q \) by \( \vec{F} = q\vec{E} \) (in the electrostatic equilibrium of charges in the conductor). So, the direction of the induced electric field is the same as the direction of \( \vec{v}\times\vec{B} \) for positive charges.

Answer:

\(+\vec{i}\)