A velocity selector uses a uniform electric field of strength 2.4×10^4 V m^-1 together with a uniform magnetic field of flux density 6.0×10^-3 T. The two fields are perpendicular to each other, and both are perpendicular to the initial velocity of a charged particle entering the…

A Level Physics 9702: Velocity Selector - Balancing Electric and Magnetic Forces (MCQ)

Step 1: Set up the force balance

For a charged particle to pass straight through a velocity selector undeflected, the electric force and the magnetic force on it must be equal in magnitude (and, as stated, opposite in direction): $F_{\text{electric}}=F_{\text{magnetic}}$ $QE=QBv$

Step 2: Solve for v

The charge $Q$ cancels from both sides, leaving: $E=Bv\implies v=\frac{E}{B}$

Step 3: Substitute the values

$v=\frac{2.4\times10^{4}}{6.0\times10^{-3}}=4.0\times10^{6}\text{ m s}^{-1}$

Recomputing as a check: $2.4\div6.0=0.4$ , and the powers of ten give $10^{4-(-3)}=10^{7}$ , so $v=0.4\times10^{7}=4.0\times10^{6}\text{ m s}^{-1}$ , the same result.

Step 4: Why the other options are wrong

B ( $1.44\times10^{2}$ ): comes from multiplying instead of dividing, $E\times B=2.4\times10^{4}\times6.0\times10^{-3}=144$ , the wrong operation entirely.
C ( $2.5\times10^{-7}$ ): comes from dividing the wrong way round, $B/E$ , instead of $E/B$ .
D ( $6.0\times10^{-3}$ ): simply restates $B$ , ignoring the electric field altogether.

Final answer

The selected speed is $\boxed{4.0\times10^{6}}\text{ m s}^{-1}$ , option A.