Question

An electron is moving with a velocity $2 \times 10^6$ m/s along positive $x$-direction in the uniform electric field of $8 \times 10^7$ V/m applied along positive $y$-direction. The magnitude and direction of a uniform magnetic field (in Tesla) that will cause the electrons to move undeviated along its original path is

• A. 40 in $-ve$ $z$-direction
• B. 40 in $+ve$ $z$-direction
• C. 4 in $+ve$ $z$-direction
• D. 4 in $-ve$ $z$-direction
• E. 8 in $+ve$ $z$-direction

Hint:

For "velocity selector" problems, remember the condition $v = E/B$. To find the direction for a negative charge, first find the direction as if it were positive, then flip the magnetic force direction.

Answer 1

The Correct Option is B

Concept:
For a charged particle to move undeviated in crossed electric and magnetic fields, the electric force ($\vec{F}_e = q\vec{E}$) and magnetic force ($\vec{F}_m = q\vec{v} \times \vec{B}$) must be equal in magnitude and opposite in direction. This means: \[ v = \frac{E}{B} \implies B = \frac{E}{v} \]

Step 1: Calculate the magnitude of the magnetic field.
Given $E = 8 \times 10^7$ V/m and $v = 2 \times 10^6$ m/s: \[ B = \frac{8 \times 10^7}{2 \times 10^6} = 40 \text{ T} \]

Step 2: Determine the direction.
[itemsep=4pt]
• Electric Force ($\vec{F}_e$): The electron is negatively charged, so the force is opposite to the electric field ($+y$). Thus, $\vec{F}_e$ is in the $-y$ direction.
• Magnetic Force ($\vec{F}_m$): Must be in the $+y$ direction to cancel $\vec{F}_e$.
• Right-Hand Rule: For a negative charge, $\vec{F}_m = -e(\vec{v} \times \vec{B})$. With $\vec{v}$ in $+x$, $\vec{B}$ must be in the $+z$ direction so that $(\hat{i} \times \hat{k}) = -\hat{j}$, and the negative charge flips it to $+\hat{j}$.