A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle between its

Question

A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle between its magnetic dipole moment and the field is 30.1∘. Find the strength B of the magnetic field that causes a torque of 1.51×10−5 N⋅m to act on the loop when a current of 2.47 mA flows in it

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Trúc Chi 5 years 2021-07-13T09:40:19+00:00 1 Answers 16 views 0

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  1. Delwyn
    0
    2021-07-13T09:41:26+00:00
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    Answer:

    B = 0.135T

    Explanation:

    To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

    \tau=NIABsin\theta   (1)

    τ: torque = 1.51*10^-5 Nm

    I: current = 2.47mA = 2.47*10^-3 A

    B: magnitude of the magnetic field

    A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

    N: turns = 181

    θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

    You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

    B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

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