Question

part a
what is the velocity of a beam of electrons that go undeflected when passing through crossed (perpendicular) electric and magnetic fields of magnitude 1.63 × 10⁴ v/m and 2.40 × 10⁻³ t, respectively?
( m_e = 9.11 × 10^{-31} ) kg, ( e = 1.60 × 10^{-19} ) c
express your answer to three significant figures and include the appropriate units.
( v = ) value units
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part b

Explanation:

Step1: Recall the condition for undeflected electrons

When electrons pass through crossed electric and magnetic fields undeflected, the electric force and magnetic force on them are equal in magnitude and opposite in direction. The electric force is \( F_E = eE \) and the magnetic force is \( F_B = evB \). Setting them equal: \( eE = evB \).

Step2: Solve for velocity \( v \)

Cancel out \( e \) from both sides of the equation \( eE = evB \), we get \( v=\frac{E}{B} \).

Step3: Substitute the given values

We know \( E = 1.63\times 10^{4}\ \text{V/m} \) and \( B = 2.40\times 10^{-3}\ \text{T} \). Plugging these into the formula: \( v=\frac{1.63\times 10^{4}}{2.40\times 10^{-3}} \).
Calculate the numerical value: \( \frac{1.63\times 10^{4}}{2.40\times 10^{-3}}=\frac{1.63}{2.40}\times 10^{4 + 3}\approx0.679167\times 10^{7}=6.79167\times 10^{6}\ \text{m/s} \). Rounding to three significant figures, we get \( v\approx6.79\times 10^{6}\ \text{m/s} \).

Answer:

The velocity of the electron beam is \( \boxed{6.79\times 10^{6}\ \text{m/s}} \)