The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. The 50-A wire

Question

The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. The 50-A wire is in the x-z plane and is 5 m from the z axis. The 40-A wire is in the y-z plane and is 4 m from the z axis. What is the magnitude of the magnetic field at the origin?

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Thu Nguyệt 4 years 2021-07-22T17:41:10+00:00 1 Answers 162 views 0

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  1. bonexptip
    0
    2021-07-22T17:43:04+00:00
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    Answer:

    The magnitude of the magnetic field at the origin is 2.56\times 10^{-6}\ T.

    Explanation:

    Given :

    50-A wire is in the x-z plane and is 5 m from the z axis.

    Also , 40-A wire is in the y-z plane and is 4 m from the z axis.

    Now , since both the wire are perpendicular to each other .

    Therefore , magnetic field are also perpendicular to each other .

    Magnetic field at origin due to wire 1 is :

    B_1=\dfrac{\mu_oI_1}{2\pi R_1}\\\\B_1=\dfrac{(50)\mu_o}{2\pi( 5)}\\\\B_1=\dfrac{5\mu_o}{\pi}

    Magnetic field at origin due to wire 2 is :

    B_2=\dfrac{\mu_oI_2}{2\pi R_2}\\\\B_2=\dfrac{(40)\mu_o}{2\pi( 4)}\\\\B_2=\dfrac{4\mu_o}{\pi}

    Now , therefore net magnetic field is :

    B=\sqrt{B_1^2+B_2^2}\\\\B=\sqrt{(\dfrac{5\mu_o}{\pi})^2+(\dfrac{4\mu_o}{\pi})^2}\\\\B=\dfrac{\sqrt{41}\mu_o}{\pi}

    Putting value of \mu_o=4\pi \times 10^{-7}\ H/m

    We get ,

    B=\sqrt{41}\times 4\times 10^{-7}\\B=2.56\times 10^{-6}\ T

    Therefore, the magnitude of the magnetic field at the origin is 2.56\times 10^{-6}\ T.

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