A wire carrying a 35.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet’s field, and

Question

A wire carrying a 35.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet’s field, and there is a 2.55 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet

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Thiên Thanh 4 years 2021-08-13T08:55:51+00:00 1 Answers 3 views 0

Answers ( )

  1. Helga
    0
    2021-08-13T08:57:03+00:00
    Reply

    Answer:

    7.65 T

    Explanation:

    From the question,

    Using,

    F = BILsinФ……………………. Equation 1

    Where F = Force, B = magnetic field strength, I = current, L = Length, Ф = Angle.

    Make B the subject of the equation

    B = F/ILsinФ……………………Equation 2

    Given: F = 2.55 N, I = 32 A, L = 3.00 cm = 0.03 m, Ф = 90° (perpendicular)

    Substitute into equation 2

    B = 2.55/(32×0.03×sin90°)

    B = 2.55/0.96

    B = 7.65 T

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