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5 September, 03:48

A 48.0-turn circular coil of radius 5.50 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.480 T. If the coil carries a current of 23.3 mA, find the magnitude of the maximum possible torque exerted on the coil.

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Answers (2)
  1. 5 September, 04:04
    0
    Given that,

    Number of turn is 48

    N=48

    Radius is 4.8cm

    r=0.048m

    Magnetic Field

    B=0.48T

    Current in coil

    i=23.3mA

    i=0.233A

    Maximum Torque?

    Maximum torque occur at angle 90°

    Torque is given as

    τ = N•I•A•B•sinθ

    Where N is number of turn = 48

    I is current in coil = 0.233A

    A is area of circular coil form

    Area of a circle is given as

    A=πr²

    A=π*0.048²

    A=0.007238m²

    B is magnetic field = 0.48T

    Maximum torque occurs at 90°

    τ = N•I•A•B•sinθ

    τ=48*0.233*0.007238*0.48*Sin90

    τ = 0.0389Nm

    This torque is large enough to exert the coil
  2. 5 September, 04:12
    0
    Given Information:

    Magnetic field = B = 0.480 T

    Current = I = 23.3 mA = 0.0233 A

    Number of turns = N = 48 turns

    Radius = r = 5.50 cm = 0.055 m

    Required Information:

    Maximum possible torque = τ = ?

    Answer:

    Maximum possible torque = 0.0051 N. m

    Explanation:

    We know that toque τ is given by

    τ = NIABsin (θ)

    Where N is the number of turns of the circular coil, I is the current flowing through the circular coil, A is the area of circular coil, B is the magnetic field induced in the circular coil.

    The area of the circular coil is

    A = πr²

    A = π (0.055) ²

    A = 0.009503 m²

    The maximum torque is possible when θ = 90°

    τ = 48*0.0233*0.009503*0.480*sin (90°)

    τ = 0.0051 N. m
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