Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

An infinitely long straight conductor carries a current of $5 \text{ A}$ . An electron is moving with a speed of $10^5 \text{ m/s}$ parallel to the conductor. The perpendicular distance between the electron and the conductor is $20 \text{ cm}$ at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

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

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

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

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

Step-by-Step Solution

Calculate Magnetic Field ( $B$ ): The magnitude of the magnetic field produced by an infinitely long straight conductor at a perpendicular distance $r$ is given by: $B = \frac{\mu_0 I}{2\pi r}$ Given $I = 5 \text{ A}$ , $r = 20 \text{ cm} = 0.2 \text{ m}$ , and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$ . $B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} = \frac{2 \times 10^{-7} \times 5}{0.2} = \frac{10^{-6}}{0.2} = 5 \times 10^{-6} \text{ T}$
Determine Angle ( $\theta$ ): The magnetic field lines around a straight wire are concentric circles. Since the electron moves parallel to the wire, its velocity vector ( $v$ ) is perpendicular to the magnetic field vector ( $B$ ) at that point (tangent to the circle). Thus, $\theta = 90^\circ$ .
Calculate Magnetic Force ( $F$ ): The force on a moving charge is given by the Lorentz force law: $F = qvB \sin\theta$ . Given electron charge $q = 1.6 \times 10^{-19} \text{ C}$ and speed $v = 10^5 \text{ m/s}$ . $F = (1.6 \times 10^{-19}) \times (10^5) \times (5 \times 10^{-6}) \times \sin(90^\circ)$ $F = 1.6 \times 5 \times 10^{-19 + 5 - 6}$ $F = 8.0 \times 10^{-20} \text{ N}$

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