A 10.0-m-long copper wire with cross-sectional area 1.25 × 10–4 m2 is bent into a square loop, and then connected to a 0.25-V battery. The l

Question

A 10.0-m-long copper wire with cross-sectional area 1.25 × 10–4 m2 is bent into a square loop, and then connected to a 0.25-V battery. The loop is then placed in a uniform magnetic field of magnitude 0.19 T. What is the maximum torque that can act on this loop? The resistivity of copper is 1.70 × 10–8 Ω·m.

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Adela 4 years 2021-07-19T23:46:49+00:00 1 Answers 12 views 0

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  1. thanhha
    0
    2021-07-19T23:47:53+00:00
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    Answer:

    Explanation:

    Length of copper is L = 10m

    Cross sectional area

    A = 1.25×10^-4m²

    Potential difference V = 0.25V

    Magnetic field B = 0.19T

    Resistivity ρ = 1.70 × 10^-8 Ω·m.

    The resistance of the wire can be calculated using

    R = ρL/A

    R = 1.70 × 10^-8 × 10/1.25×10^-4

    R = 1.36×10^-3 ohms

    Using ohms law

    v= IR

    Then, I = V/R

    I = 0.25/1.36×10^-3

    I = 183.82 Amps

    Maximum torque is give as

    τ = NiAB

    If the wire form a square

    Then each side is L/4 = 10/4 = 2.5m

    Then, Area of a square is L²

    A = 2.5² = 6.25m²

    Number of turn is assume to be 1

    N=1

    Therefore

    τ = 1×183.82×6.25×0.19

    τ = 218.29 Nm

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