A circular loop of radius 15 cm carries a current of 11 A. A flat coil of radius 0.79 cm, having 66 turns and a current of 1.9 A, is concent

Question

A circular loop of radius 15 cm carries a current of 11 A. A flat coil of radius 0.79 cm, having 66 turns and a current of 1.9 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop’s magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop

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Sigridomena 1 year 2021-08-31T21:32:11+00:00 1 Answers 3 views 0

Answers ( )

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    2021-08-31T21:34:09+00:00

    Answer:

    [tex]4.61\times 10^{-5}\ \text{T}[/tex]

    [tex]1.05\times 10^{-6}\ \text{Nm}[/tex]

    Explanation:

    [tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]

    [tex]r_l[/tex] = Radius of loop = 15 cm

    [tex]I_l[/tex] = Current in loop = 11 A

    [tex]r_c[/tex] = Radius of coil = 0.76 cm

    N = Number of turns of coil = 66

    [tex]I_c[/tex] = Current in coil = 1.9 A

    Magnetic field is given by

    [tex]B=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}\times 11}{2\times 0.15}\\\Rightarrow B=4.61\times 10^{-5}\ \text{T}[/tex]

    Magnitude of magnetic field produced by the loop at its center is [tex]4.61\times 10^{-5}\ \text{T}[/tex].

    Torque is given by

    [tex]\tau=BI_c\pi r_c^2N\sin90^{\circ}\\\Rightarrow \tau=4.61\times 10^{-5}\times 1.9\times \pi\times (0.76\times 10^{-2})^2\times 66\sin90^{\circ}\\\Rightarrow \tau=1.05\times 10^{-6}\ \text{Nm}[/tex]

    Magnitude of torque on the coil due to the loop is [tex]1.05\times 10^{-6}\ \text{Nm}[/tex]

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