slader A long, straight wire carries a current of 5.20 AA. An electron is traveling in the vicinity of the wire. At the instant when the ele

Question

slader A long, straight wire carries a current of 5.20 AA. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cmcm from the wire and traveling with a speed of 6.20×104 m/sm/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron?

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RobertKer 2 weeks 2021-08-29T19:02:02+00:00 1 Answers 0 views 0

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    2021-08-29T19:03:16+00:00

    Answer:

    2.34\cdot 10^{-19}N

    Explanation:

    The strength of the magnetic field produced by a current-carrying wire is:

    B=\frac{\mu_0 I}{2\pi r}

    where:

    \mu_0 is the vacuum permeability

    I is the current in the wire

    r is the distance from the wire

    In this case,

    I = 5.20 A

    r = 4.40 cm = 0.044 m

    Therefore,

    B=\frac{(4\pi \cdot 10^{-7})(5.20)}{2\pi (0.044)}=2.36\cdot 10^{-5} T

    The direction of the field is along concentric circles centered in the wire.

    The force exerted by a magnetic field on a moving charged particle is

    F=qvB sin \theta

    where

    q is the charge

    v is the velocity of the particle

    \theta is the direction between v and B

    In this problem:

    q=1.6\cdot 10^{-19}C is the magnitude of the charge of the particle

    v=6.20\cdot 10^4 m/s is the velocity

    \theta=90^{\circ}, because the electron is travelling towards the wire, so perpendicular to the lines of the magnetic field

    Therefore, the magnitude of the force is:

    F=(1.6\cdot 10^{-19})(2.36\cdot 10^{-5})(6.20\cdot 10^4)(sin 90^{\circ})=2.34\cdot 10^{-19}N

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