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30 November, 08:36

A 7.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 Ω. Pulling the wire at a steady speed of 4.00 m/s causes 4.30 W of power to be dissipated in the circuit.

How big is the pulling force?

What is the strength of the magnetic field?

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Answers (1)
  1. 30 November, 09:10
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    (a) 1.075 N

    (b) 4.254 T

    Explanation:

    (a)

    From the question,

    Power = Force*Velocity.

    P = Fv ... Equation 1

    Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.

    make F the subject of the equation

    F = P/v ... Equation 2

    Given: P = 4.30 W, v = 4.0 m/s.

    Substitute into equation 2

    F = 4.30/4

    F = 1.075 N.

    (b)

    Applying,

    F = BILsinФ ... Equation 2

    Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.

    make B the subject of the equation

    B = F/ILsinФ ... Equation 3

    But,

    P = I²R ... Equation 4

    Where R = resistance of the wire.

    make I the subject of the equation

    I = √ (P/R) ... Equation 5

    Given: P = 4.30 W, R = 0.330 Ω

    Substitute into equation 5

    I = √ (4.3/0.33)

    I = √13.03

    I = 3.61 A.

    Also given: L = 7 cm = 0.07 m, Ф = 90°

    Substitute into equation 3

    B = 1.075 / (0.07*3.61*sin90)

    B = 1.075/0.2527

    B = 4.254 T.
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