Question

An electric field of \(1000 \, V/m\) and a perpendicular magnetic field act on a moving electron to produce no net force. If the speed of the electron is \(2.5 \, km/s\), then the magnitude of the magnetic field is:

• A. 0.4 T
• B. 0.25 T
• C. 0.6 T
• D. 0.75 T

Hint:

For zero net force, velocity must satisfy \(v = E/B\) in perpendicular E and B fields.

Answer 1

The Correct Option is A

Step 1: Identify the two forces on the electron directly, without vector cross-product notation.
The electron feels an electric force of size $qE$ along the field, and a magnetic force of size $qvB$ perpendicular to its motion, set up here to oppose the electric force.

Step 2: Use the condition of zero net force.
No net force means these two forces match in size: \[ qE = qvB \] The charge cancels, leaving \[ E = vB \] which is the working condition of a velocity selector: only a charge moving at exactly this speed passes through undeflected.

Step 3: Convert the given speed to SI units.
\[ v = 2.5\ \text{km/s} = 2500\ \text{m/s} \]
Step 4: Solve for $B$.
\[ B = \frac{E}{v} = \frac{1000}{2500} = 0.4\ \text{T} \]
Step 5: Why the other options are wrong.
$0.25$ T, $0.6$ T and $0.75$ T would each need a different speed or field than the ones given; only $1000/2500$ reduces to $0.4$.

Final Answer:
\[ \boxed{0.4\ \text{T}} \]