Question

module 8a
problem 27.02
1 of 16
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_{\text{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 = 6.79×10^{6} \frac{\text{m}}{\text{s}} )
submit previous answers
correct
part b
what is the radius of the electron orbit if the electric field is turned off?
express your answer to three significant figures and include the appropriate units.
( r = ) value units

Explanation:

Step1: Recall the formula for radius in magnetic field

When the electric field is off, the electron moves in a circular path due to the magnetic force, which provides the centripetal force. The formula is \( F_{magnetic} = F_{centripetal} \), i.e., \( evB=\frac{mv^{2}}{r} \). Solving for \( r \), we get \( r = \frac{mv}{eB} \).

Step2: Substitute the known values

We know \( m_e = 9.11\times 10^{-31}\space kg \), \( v = 6.79\times 10^{6}\space m/s \) (from part A), \( e = 1.60\times 10^{-19}\space C \), and \( B = 2.40\times 10^{-3}\space T \).
Substitute these values into the formula:

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Wait, let's recalculate the numerator and denominator more accurately:
Numerator: \( 9.11\times 10^{-31}\times6.79\times 10^{6}=9.11\times6.79\times 10^{-25}\approx61.8569\times 10^{-25} = 6.18569\times 10^{-24} \)
Denominator: \( 1.60\times 10^{-19}\times2.40\times 10^{-3}=3.84\times 10^{-22} \)
Now divide: \( \frac{6.18569\times 10^{-24}}{3.84\times 10^{-22}}=\frac{6.18569}{3.84}\times 10^{-24 + 22}\approx1.611\times 10^{-2}\space m = 0.0161\space m = 1.61\times 10^{-2}\space m \) or \( 1.61\space cm \) (but in meters, it's \( 0.0161\space m \) or \( 1.61\times 10^{-2}\space m \))
Wait, let's check the calculation again:
\( 9.11\times10^{-31} \times 6.79\times10^{6}=9.11\times6.79\times10^{-25}\). \( 9.11\times6.79 = 9\times6.79+0.11\times6.79 = 61.11 + 0.7469 = 61.8569 \). So \( 61.8569\times10^{-25}=6.18569\times10^{-24} \)
\( 1.60\times10^{-19}\times2.40\times10^{-3}=3.84\times10^{-22} \)
Now, \( \frac{6.18569\times10^{-24}}{3.84\times10^{-22}}=\frac{6.18569}{3.84}\times10^{-24 + 22}=\frac{6.18569}{3.84}\times10^{-2}\)
\( 6.18569\div3.84\approx1.611 \), so \( 1.611\times10^{-2}\space m = 0.0161\space m \) (to three significant figures, \( 0.0161\space m \) or \( 1.61\times 10^{-2}\space m \))

Answer:

The radius of the electron orbit is \(\boldsymbol{1.61\times 10^{-2}\space m}\) (or \(0.0161\space m\))