A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from the w

Question

A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from
the wire.

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bonexptip 3 weeks 2021-09-03T17:56:42+00:00 1 Answers 0 views 0

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    2021-09-03T17:58:41+00:00

    Answer:

    1.67\cdot 10^{-5} T

    Explanation:

    The magnitude of the magnetic field produced by a current-carrying wire is given by the equation:

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

    where:

    \mu_0=4\pi \cdot 10^{-7} H/m is the vaacuum permeability

    I is the current in the wire

    r is the distance from the wire

    The direction of the magnetic field lines is tangential to concentric circles around the wire.

    In this problem, we have:

    I=37.9 A is the current in the wire

    r=45.5 cm = 0.455 m is the distance  from the wire

    Solving for B, we find the magnitude of the magnetic field:

    B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi \cdot 10^{-7})(37.9)}{2\pi (0.455)}=1.67\cdot 10^{-5} T

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