A wire has a length of 7.00 \times 10^{-2} m and is used to make a circular coil of one turn. There is a current of 4.30 A in the wire. In t

Question

A wire has a length of 7.00 \times 10^{-2} m and is used to make a circular coil of one turn. There is a current of 4.30 A in the wire. In the presence of a 2.50-T magnetic field, what is the maximum torque that this coil can experience

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niczorrrr 4 years 2021-08-22T21:15:52+00:00 1 Answers 26 views 0

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  1. ngockhue
    0
    2021-08-22T21:17:02+00:00
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    Answer:

    Maximum torque that coil experience is 4.19 x 10⁻³ N-m

    Explanation:

    Given :

    Length of wire, L = 7 x 10⁻² m

    Current flowing through the circular coil, I = 4.30 A

    Magnetic field applied, B = 2.50 T

    Number of turns, N = 1

    The relation to determine torque experienced by the circular coil due to magnetic field:

    τ = NIAB sinθ   ….(1)

    Here A is the area of the circular coil and θ is the angle between circular coil and magnetic field.

    Consider the radius of the circular coil be r.

    But, length of wire = Circumference of the circular coil

    So, L = 2πr

    r = L/2π    ….(2)

    Area of circular coil, A = πr²

    Substitute equation (2) in the above equation.

    A=\frac{L^{2} }{4\pi }

    Substitute the above equation in equation (1)

    \tau=\frac{NIBL^{2} \sin \theta}{4\pi }

    The torque will be maximum when sin θ = 1, that is, θ = 90⁰. Thus, the above equation becomes:

    \tau=\frac{NIBL^{2}}{4\pi }

    Substitute the suitable values in the above equation.

    \tau=\frac{1\times4.30\times2.50\times(7\times10^{-2}) ^{2}}{4\pi }

    τ = 4.19 x 10⁻³ N-m

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