An electron is moving along the positive $\mathrm{x}$-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then

A. The electron will experience magnetic force along positive y-axis

B. The electron will experience magnetic force along negative y-axis

C. The electron will not experience any force in magnetic field

D. The electron will continue to move along the positive $\mathrm{x}$-axis

E. The electron will move along circular path in magnetic field

Choose the correct answer from the options given below:

The Lorentz force equation is given as: <br/><br/> $\vec{F} = -q(\vec{v} \times \vec{B})$ <br/><br/> The electron is moving along the positive x-axis, so its velocity vector is $\vec{v} = v_x \hat{i}$. The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is $\vec{B} = -B_z \hat{k}$. <br/><br/> Now, we can calculate the cross product of the velocity and magnetic field vectors: <br/><br/> $\vec{v} \times \vec{B} = (v_x \hat{i}) \times (-B_z \hat{k})$ <br/><br/> Using the cross product properties, we get: <br/><br/> $\vec{v} \times \vec{B} = -v_x B_z (\hat{i} \times \hat{k})$ <br/><br/> The cross product of $\hat{i}$ and $\hat{k}$ is $-\hat{j}$, so: <br/><br/> $\vec{v} \times \vec{B} = -v_x B_z (-\hat{j}) = v_x B_z \hat{j}$ <br/><br/> Since the electron has a negative charge, the magnetic force will be in the opposite direction: <br/><br/> $\vec{F} = -(-e)(v_x B_z \hat{j}) = e(v_x B_z \hat{j})$ <br/><br/> As a result, the electron will experience a magnetic force along the negative y-axis. <br/><br/> Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path. In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other. As a result, the electron will move along a circular path in the magnetic field. <br/><br/> Hence, the correct answer is: <br/><br/> (C) B and E only