An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵m/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.

The problem involves an infinitely long straight conductor carrying a current $I\; =\; 5 A$. An electron moves parallel to the wire with speed $v\; =\; 1 \times 105 m/s$ at a perpendicular distance $r\; =\; 20 cm\; =\; 0.20 m$ from the wire, as illustrated in the provided diagram.

First, compute the magnetic field produced by the current at the location of the electron. For a long straight conductor the magnetic field magnitude is given by Ampère’s law:

$B\; =\; \backslash frac\{\backslash mu\_0\; I\}\{2\backslash pi\; r\}$

where $\backslash mu\_0\; =\; 4\backslash pi\; \backslash times\; 10^\{-7\}\backslash ,\; \backslash text\{T·m/A\}$ is the permeability of free space.

Substituting the values:

$B\; =\; \backslash frac\{(4\backslash pi\; \backslash times\; 10^\{-7\}\backslash ,\backslash text\{T·m/A\})(5\backslash ,\backslash text\{A\})\}\{2\backslash pi\; (0.20\backslash ,\backslash text\{m\})\}$

Cancel the factor π and simplify:

$B\; =\; \backslash frac\{4\; \backslash times\; 5\; \backslash times\; 10^\{-7\}\}\{2\; \backslash times\; 0.20\}$

$B\; =\; \backslash frac\{20\; \backslash times\; 10^\{-7\}\}\{0.40\}$

$B\; =\; 5\; \backslash times\; 10^\{-6\}\backslash ,\backslash text\{T\}$ (or 5 µT).

The magnetic force on a moving charge is

$F\; =\; |q|\backslash ,v\backslash ,B\backslash ,\backslash sin\backslash theta$

where $\backslash theta$ is the angle between the velocity vector and the magnetic field. In this configuration the velocity is parallel to the wire, while the magnetic field circles the wire, making the angle 90°; thus $\backslash sin\backslash theta\; =\; 1$.

For an electron, the charge magnitude is $|q|\; =\; e\; =\; 1.6\; \backslash times\; 10^\{-19\}\backslash ,\backslash text\{C\}$. Hence:

$F\; =\; (1.6\; \backslash times\; 10^\{-19\}\backslash ,\backslash text\{C\})\backslash ,(1\; \backslash times\; 10^\{5\}\backslash ,\backslash text\{m/s\})\backslash ,(5\; \backslash times\; 10^\{-6\}\backslash ,\backslash text\{T\})$

Calculate the product of $v$ and $B$ first:

$vB\; =\; (1\; \backslash times\; 10^\{5\})(5\; \backslash times\; 10^\{-6\})\; =\; 0.5$

Then multiply by the charge:

$F\; =\; (1.6\; \backslash times\; 10^\{-19\})\; \backslash times\; 0.5\; =\; 0.8\; \backslash times\; 10^\{-19\}\backslash ,\backslash text\{N\}$

Expressing the result in standard scientific notation:

$F\; =\; 8\; \backslash times\; 10^\{-20\}\backslash ,\backslash text\{N\}$

Therefore, the magnitude of the force experienced by the electron at that instant is 8 × 10⁻²⁰ N, which matches the given correct option.