Physics 2

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A wire has a length of 4.52 x 10-2 m and is used to make a circular coil of one turn. There is a current of 7.94 A in the wire. In the presence of a 8.04-T magnetic field, what is the maximum torque that this coil can experience?

  • Physics 2 -

    A 45 turn coil of radius 5.1 cm rotates in a uniform magnetic field having a magnitude of 0.49 T. If the coil carries a current of 40 mA, find the magnitude of the maximum torque exerted on the coil.

  • Physics 2 -

    Torque = (the number of turns)(current)(Area)((magnetic field)(sin(angle)))

    Area = π × r^2
    the angle is the angle between the normal to the plane of the coil and the magnetic field.

    Torque = (45)(40)(π(5.1)^2)(0.49 sin(90))
    Maximum torque occurs when the magnetic force is perpendicular to the normal (which is sin(90)= 1

    = (45)(40)(π(5.1)^2)(0.49)
    = 72070.71158 N*m

  • Physics 2 -

    @ Megan
    ----------------------

    Torque = (# of turns)(current)(Area)((magnetic field)(sin(angle)))

    Area = (circumference)^2 / (4π)
    circumference is distance around the edge of a circle.

    Torque = (1)(7.94)(((4.52 x 10^-2)^2)/(4π))(8.04 sin(90))

    =(7.94)(((4.52 x 10^-2)^2)/(4π))(8.04)

    =(7.94)((.0452)^2/(4π))(8.04)

    = .0103787 N*m

    WHEN YOU CALCULATE AREA ON YOUR CALCULATOR MAKE SURE TO PUT BRACKETS AROUND THE 4π.
    ****THIS --> (.0452)^2/(4π)
    NOT THIS --> (.0452)^2/4π

    IF YOU DON'T ADD BRACKETS THEN IT WILL NOT GIVE YOU THE CORRECT AREA.

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