Q 12-04-011NEETNEET 2021Top questionMedium
An infinitely long straight conductor carries a current of $5$ A as shown. An electron is moving with a speed of $10^5$ 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.
Answer: (A) $8 \times 10^{-20}$ N
Field of the wire at $20$ cm:
$$B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 5}{0.2} = 5 \times 10^{-6}\ \text{T}$$
The electron moves parallel to the wire, perpendicular to $B$:
$$F = evB = 1.6 \times 10^{-19} \times 10^5 \times 5 \times 10^{-6} = 8 \times 10^{-20}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics