Question

A long straight wire carries a current of 10a. An electron travels with a speed of 5 x 10^6 m/a parallel to the wire 0.1m away from it. The force exerted by maneuver field of the current electron is?

Answer 1

1.6*10^-18 N so guys as the distance is 10cm = 0.1m from the first wire it slows the force or speed of electron

Answer 2

Answer:

The force exerted by the magnetic field of the current electron is $F = 1.6\times10^{-17}\ N$[/tex].

Explanation:

Given that,

Current I = 10 A

Speed $v =5\times10^6\ m/s$

Length l = 0.1 m

We know that,

The force exerted by the magnetic field is

$F = qv\times B$

The magnetic field is

$B = \dfrac{\mu_{0}I}{2\pi\times d}$

$B = \dfrac{4\pi\times10^{-7}\times 10}{2\pi\times0.1}$

$B = 2\times 10^{-6} T$

The force will be

$F = 1.6\times10^{-19}\times5\times10^{6}\times20\times10^{-6}$

$F = 1.6\times10^{-17}\ N$

Hence, The force exerted by the magnetic field of the current electron is $F = 1.6\times10^{-17}\ N$.