Ask Question
4 January, 14:20

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.

+5
Answers (1)
  1. 4 January, 14:39
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers