If a charge ion goes through a combined electric field E and magnetic field B, the resultant emergent velocity of the ion is

Analysis

The question describes a situation where a charged particle moves through a region containing both an electric field ($E$) and a magnetic field ($B$) acting perpendicularly to each other. This is the principle behind a Velocity Selector.

Concept

When a charged particle (with charge $q$ and velocity $v$) enters a region with both fields, it experiences two forces:

1. Electric Force ($F_E$): Given by $F_E = qE$.
2. Magnetic Force ($F_B$): Given by $F_B = qvB$ (assuming the velocity is perpendicular to the magnetic field).

For the ion to pass through the fields without being deflected (the "emergent velocity"), the electric force must be equal and opposite to the magnetic force. This means the net force on the particle is zero.

Step-by-Step Solution

1. Set the forces equal to each other: $$F_E = F_B$$ $$qE = qvB$$

2. Simplify the equation: Notice that the charge ($q$) appears on both sides, so it cancels out: $$E = vB$$

3. Solve for velocity ($v$): To isolate $v$, divide both sides by $B$: $$v = \frac{E}{B}$$

Conclusion

The resultant emergent velocity of the ion is the ratio of the electric field strength to the magnetic field strength. This allows the particle to travel in a straight line through the fields.

Correct Option: A

Exam Tip

Remember the mnemonic "E over B" for velocity selectors. If you ever forget the formula, just remember that the units must work out: Electric field is in $V/m$ and Magnetic field is in $Tesla$ ($N/Am$). Dividing them gives you $m/s$, which is the unit for velocity!