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NEET PHYSICSGeneralMedium

Question

An infinitely long straight conductor carries a current of 5 A5\text{ A} as shown. An electron is moving with a speed of 105 m/s10^5\text{ m/s} parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm20\text{ cm} at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

A

4×10−20 N4 \times 10^{-20}\text{ N}

B

8π×10−20 N8\pi \times 10^{-20}\text{ N}

C

4π×10−20 N4\pi \times 10^{-20}\text{ N}

D

8×10−20 N8 \times 10^{-20}\text{ N}

Step-by-Step Solution

The magnetic field BB at distance rr from a long wire is B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}. The force on the electron is F=qvBsin⁡(θ)F = qvB \sin(\theta). Here q=1.6×10−19 Cq = 1.6 \times 10^{-19}\text{ C}, v=105 m/sv = 10^5\text{ m/s}, I=5 AI = 5\text{ A}, r=0.2 mr = 0.2\text{ m}. B=4π×10−7×52π×0.2=5×10−6 TB = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} = 5 \times 10^{-6}\text{ T}. F=(1.6×10−19)×105×(5×10−6)=8×10−20 NF = (1.6 \times 10^{-19}) \times 10^5 \times (5 \times 10^{-6}) = 8 \times 10^{-20}\text{ N}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSinfinitelystraightconductorcarriescurrent

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