A rod of m= 1.3 kg rests on two parallel rails that are L = 0.42 m apart. The rod carries a current going between the rails (bottom to top i

Question

A rod of m= 1.3 kg rests on two parallel rails that are L = 0.42 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 2.6 A. A uniform magnetic field of magnitude B = 0.35 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d=1.25 m. Ignore the friction on the rails. † † † Ē I Otheexpertta.com A Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails.Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times.

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Thu Giang 4 years 2021-08-21T05:02:06+00:00 1 Answers 25 views 0

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  1. yenoanh
    0
    2021-08-21T05:03:40+00:00
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    Answer:

    The final speed of the rod is 0.86 m/s.

    Explanation:

    Given that,

    Mass of rod = 1.3 kg

    Distance between rail= 0.42 m

    Current = 2.6 A

    Magnetic field = 0.35 T

    Distance = 1.25 m

    We need to calculate the acceleration

    Using formula of magnetic force

    F= Bil

    ma=Bil

    a=\dfrac{Bil}{m}

    Put the value into the formula

    a=\dfrac{0.35\times2.6\times0.42}{1.3}

    a=0.294\ m/s^2

    We need to calculate the final speed of the rod

    Using equation of motion

    v^2-u^2=2as

    Put the value in the equation

    v^2=2\times0.294\times1.25

    v=0.86\ m/s

    Hence, The final speed of the rod is 0.86 m/s.

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