An infinitely long straight conductor carries a current of $5 A$ as shown. An electron is moving with a speed of $10^{5} m s^{- 1}$ 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.

Question

Aicharya · Accepted Answer

Magnetic Field due to a current carrying wire at $20 cm$ in terms of current $I$ is given by,

$B = \frac{μ_{0} I}{2 π d}$

$B = \frac{2 \times 10^{- 7} \times 5}{20 \times 10^{- 2}}$

$B = 0.5 \times 10^{- 5} T$

Now,

Force on charge particle $= qvB$

$F_{m} = e V B$

$F_{m} = 1.6 \times 10^{- 19} \times 10^{5} \times 0.5 \times 10^{- 5}$

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

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An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed…

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