4.21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?

(Ignore the mass of the wires.) g = 9.8 m s–2.

0 2 Views | Posted 5 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    4.21 Length of the rod. l= 0.45 m

    Mass suspended by the wire, m = 60 g = 60 × 10 - 3 kg

    Acceleration due to gravity, g = 9.8 m/ s 2

    Current, I = 5 A

    To achieve zero tension, the magnetic field = weight of the wire

    BIl = mg or

    B = m g I l = 60 × 10 - 3 × 9.8 5 × 0.45 = 0.26 T

    The magnetic field should be set up such that it gives an upward magnetic force.

    If the direction of current is reversed, then the magnetic force will act downwards and total tension in the wire will be

    mg + BIl = 60 × 10 - 3 × 9.8 + 0.26 × 5 × 0.45 = 1.173 T

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