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A long, straight wire carries a current of 4.23 A. An electron travels at 47100 m/s parallel to the wire, 61.1 cm from the wire. The permeability of free space is 1.25664 × 10−6 N/A 2 and the charge on an electron is 1.6 × 10−19 C. What force does the magnetic field of the current exert on the moving electron? Answer in units of N.

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Answer:

[tex]F_e = 1.043*10^{-20} N[/tex]

Explanation:

To give a proper solution to this problem we need to apply the concept about Force on electron, which is given by,

[tex]F_e = BeV[/tex]

Where,

[tex]F_e[/tex]= Perpendicular Force to the direction of magnetic field

B = Magnetic field

V = Velocity (Perpendicular also)

We know that B is given by

[tex]B = \frac{\mu_0I}{2\pi d}[/tex]

Where [tex]\mu[/tex] is the permeability constant I is the current and d the distance

Replacing in the Force equation,

[tex]F_e = \frac{\mu_0I}{2\pi d} eV[/tex]

[tex]F_e = \frac{(4\pi*10^{-7})(4.23)} {2\pi 0.611}*1.6*10^{-19}*47100[/tex]

[tex]F_e = 1.043*10^{-20} N[/tex]

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