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NEET PHYSICSMedium

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}.

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