MEC O7

For an electron to move undeviated in combined electric and magnetic fields, the net Lorentz force must be zero:

$$ \vec{F} = -e [\vec{E} + (\vec{v} \times \vec{B})] = 0 \implies \vec{E} = -(\vec{v} \times \vec{B}) = \vec{B} \times \vec{v} $$

Analyzing the Two Cases

Case 1: Velocity $\vec{v}_1 = v\hat{i}$.

$$ \vec{E} = \vec{B} \times (v\hat{i}) $$

Case 2: Velocity $\vec{v}_2 = v\hat{j}$.

$$ \vec{E} = \vec{B} \times (v\hat{j}) $$

Equating the expression for $\vec{E}$:

$$ \vec{B} \times v\hat{i} = \vec{B} \times v\hat{j} $$ $$ \vec{B} \times (\hat{i} – \hat{j}) = 0 $$

For the cross product to be zero, the magnetic field $\vec{B}$ must be parallel to the vector $(\hat{i} – \hat{j})$.

Conclusions

• Direction of B (Option d): Since $\vec{B}$ is parallel to $\hat{i} – \hat{j}$, it makes an angle of $135^\circ$ (or $-45^\circ$) with the positive x-axis. Option (d) is correct.
• Direction of E (Option c): Using $\vec{E} = \vec{B} \times \vec{v}_1$, and knowing $\vec{B}$ lies in the x-y plane: $$ \vec{E} \propto (\hat{i} – \hat{j}) \times \hat{i} = -(\hat{j} \times \hat{i}) = \hat{k} $$ The electric field is along the z-axis, which is perpendicular to the x-y plane. Option (c) is correct.
• Magnitude Comparison (Option a): The angle $\theta$ between $\vec{B}$ (along $135^\circ$) and $\vec{v}_1$ (along $0^\circ$) is $135^\circ$. $$ E = v B \sin(135^\circ) = v B \frac{1}{\sqrt{2}} $$ $$ E = \frac{vB}{\sqrt{2}} $$ Since $\sqrt{2} > 1$, it follows that $E < vB$ (or $|\vec{E}| < v|\vec{B}|$). Option (a) is correct.