Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the current in t

Question

Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the current in the other is 3.15 A. Find the magnitude of the force per unit length that one wire exerts on the other.

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Thiên Di 4 weeks 2021-08-30T20:43:46+00:00 2 Answers 0 views 0

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    0
    2021-08-30T20:45:44+00:00

    Answer:

    Therefore,

    The magnitude of the force per unit length that one wire exerts on the other is

    \dfrac{F}{l}=2.79\times 10^{-5}\ N/m

    Explanation:

    Given:

    Two long, parallel wires separated by a distance,

    d = 3.50 cm = 0.035 meter

    Currents,

    I_{1}=1.55\ A\\I_{2}=3.15\ A

    To Find:

    Magnitude of the force per unit length that one wire exerts on the other,

    \dfrac{F}{l}=?

    Solution:

    Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

    \dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

    where,

    \mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}

    Substituting the values we get

    \dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}

    \dfrac{F}{l}=2.79\times 10^{-5}\ N/m

    Therefore,

    The magnitude of the force per unit length that one wire exerts on the other is

    \dfrac{F}{l}=2.79\times 10^{-5}\ N/m

    0
    2021-08-30T20:45:45+00:00

    Answer:

    The force per unit length is 2.79 \times 10^{-5}~N.

    Explanation:

    As depicted in the figure, the current (I_{1}) through wire 1 is 1.55 A upwards and the current (I_{2}) through wire 2 is 3.15 A downwards. The wires are 3.50 cm apart.

    From Biot-Savart’s law, we know that the magnetic field (B) due to a current carrying conductor carrying a current I at a distance r from the conductor is given by

    \vec{B} = \dfrac{\mu_{0}I}{4 \pi} \oint \dfrac{\vec{dl} \times \hat{r}}{r^{2}}

    According to the figure, the magnetic field (B) at 3.5 cm right to wire 1 due to it can be written as

    B = \dfrac{\mu_{0}I_{1}}{2\pi \times 0.035~m} = \dfrac{4 \pi \times10^{-7} N~A^{-2}\times 1.55~A}{2 \pi \times 0.035~m} = 88.57 \times 10^{-7} T

    andthe direction will be perpendicular to the screen and towards the screen (as depicted by the \bigotimes symbol in the figure).

    Also we know that the force ‘F‘ experience by  a current carrying conductor of length ‘L‘ having current ‘I‘ and placed in a magnetic field ‘B‘ is given by

    F = I~L~B

    So the force per unit length (F_{21}) experienced by wire 2 due to the magnetic field due to wire 1 is given by

    F_{21} = \dfrac{F}{L} = I_{2} \times B = 3.15~A \times 88.57 \times 10^{-7}~T = 2.79 \times 10^{-5} N

    and the direction of the force will be right to wire 2.

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