An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 105 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

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$8 \times 10^{- 20} N$

$8 π \times 10^{- 20} N$

$4 \times 10^{- 20} N$

$4 π \times 10^{- 20} N$

answer is C.

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Detailed Solution

$\begin{aligned} B = \frac{μ_{0} t}{2 π R} \\ F = B V q \sin θ \\ θ = 90^{\circ} \\ F = B V_{q} \\ F = \frac{μ_{0} I}{2 π R} \times V \times e = \frac{2 \times 10^{- 7} \times 5}{20 \times 10^{- 2}} \times 10^{5} \times 1.6 \times \\ 10^{- 19} \end{aligned}$

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