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31 May, 08:34

In a classroom demonstration, two long parallel wires are separated by a distance of 2.25 cm. One wire carries a current of 1.30 A while the other carries a current of 3.15 A. The currents are in the same direction. (a) What is the magnitude of the force per unit length (in N/m) that one wire exerts on the other

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  1. 31 May, 09:07
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    Given that,

    Current in wire are 1.3A and 3.15A

    Distance between wire is d = 2.25cm

    d = 2.25/100 = 0.025m

    Force per unit length F/l?

    Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2).

    The field due to I1 at a distance r is given to be

    B1 = μo• I1 / 2πr

    This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by

    F=ILBsinθ

    with sinθ=1:

    F2=I2 • L •B1

    By Newton's third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2. (Note that F1=-F2.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for B1 into the last equation and rearranging terms gives

    F/l = μo• I1• I2 / 2πr

    Where μo is constant

    μo = 4π*10^7 Tm/A

    Then,

    F/l = μo• I1• I2 / 2πr

    F/l = 4π * 10^-7 * 1.3*3.15 / (2π*0.025)

    F/l = 3.276*10^-5 N/m

    the magnitude of the force per unit length that one wire exerts on the other is 3.276*10^-5 N/m
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