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A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cm from the wire and traveling at a speed of 6.20 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron

Sagot :

Answer:

Explanation:

Magnetic field due to current at a distance of 4.4 cm

B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻²           [ B = 10⁻⁷ x 2i / r = ]

= 2.36 x 10⁻⁵ T.

Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .

Force = 2.36 x 10⁻⁵  x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴

= 23.41 x 10⁻²⁰ N .

This force will be perpendicular to the direction of current .

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