To find the magnetic force acting on the electron, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:
\[ F = qvB \sin(\theta) \]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle (for an electron, \( q = -1.6 \times 10^{-19} \) C),
- \( v \) is the velocity of the particle (in this case, \( v = 5.5 \times 10^4 \) m/s),
- \( B \) is the magnetic field strength (in this case, \( B = 2 \) T),
- \( \theta \) is the angle between the velocity vector and the magnetic field vector. Since the electron is moving at right angles to the magnetic field, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1\).
Now substituting the values into the formula:
\[ F = (1.6 \times 10^{-19} , \text{C}) \cdot (5.5 \times 10^{4} , \text{m/s}) \cdot (2 , \text{T}) \cdot \sin(90^\circ) \]
Calculating:
\[ F = (1.6 \times 10^{-19}) \cdot (5.5 \times 10^{4}) \cdot 2 \]
\[ F = (1.6 \times 5.5 \times 2) \times 10^{-19} \cdot 10^{4} \]
\[ F = (17.6) \times 10^{-15} \text{N} = 1.76 \times 10^{-14} \text{N} \]
This result can be approximated to:
\[ F \approx 1.8 \times 10^{-14} , \text{N} \]
Thus, the correct answer is:
1.8 x 10-14 N
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