A long straight wire is carrying a current of 20 A. An electron is moving with a speed of $10^7$ m/s parallel to the wire. The force experienced by the electron when it is 2 cm away from the wire is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $3.2 × 10^{-16} N$
Given:
Current: $I = 20 \, \text{A}$
Electron speed: $v = 10^7 \, \text{m/s}$ (parallel to wire)
Distance from wire: $r = 2 \, \text{cm} = 0.02 \, \text{m}$
Magnetic field due to long straight wire at distance $r$:
$B = \frac{\mu_0 I}{2 \pi r}$
$\mu_0 = 4 \pi \times 10^{-7} \, \text{H/m}$
$B = \frac{4 \pi \times 10^{-7} \cdot 20}{2 \pi \cdot 0.02} = \frac{8 \pi \times 10^{-6}}{0.04 \pi} = 2 \times 10^{-4} \, \text{T}$
Force on moving electron: $\vec{F} = q (\vec{v} \times \vec{B})$
Since velocity is parallel to the wire and magnetic field is circular around wire, $\vec{v} \perp \vec{B}$, so $F = e v B$
Electron charge: $e = 1.6 \times 10^{-19} \, \text{C}$
$F = (1.6 \times 10^{-19}) (10^7) (2 \times 10^{-4}) = 3.2 \times 10^{-16} \, \text{N}$
Answer: $F = 3.2 \times 10^{-16} \, \text{N}$