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Physics Question on Magnetic Force

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

A

4×10−20N4 \times 10^{-20} N

B

8π×10−20N8 \pi \times 10^{-20} N

C

4π×10−20N4 \pi \times 10^{-20} N

D

8×10−20N8 \times 10^{-20} N

Answer

8×10−20N8 \times 10^{-20} N

Explanation

Solution

B=μ0ii2πrB=\frac{\mu_{0} i^{i}}{2 \pi r}
F=evB=ev×μ0i2πrF=e v B=\frac{e v \times \mu_{0} i}{2 \pi r}
=(1.6×10−19)×105×4π×10−7×52π×0.2 N=\frac{\left(1.6 \times 10^{-19}\right) \times 10^{5} \times 4 \pi \times 10^{-7} \times 5}{2 \pi \times 0.2} \,N
=1.6×2×52×10−20 N=\frac{1.6 \times 2 \times 5}{2} \times 10^{-20} \,N
⇒F=8×10−20 N\Rightarrow F=8 \times 10^{-20} \,N