A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical. The magnitude of the magnetic f

Question

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical. The magnitude of the magnetic field B changes in time according to the equation B(t)=(3.75 T)+(2.75 Ts)t+(−7.05 Ts2)t2 If the radius of the wire loop is 0.210 m, find the magnitude  of the induced emf in the loop when t=5.63 s.

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Diễm Kiều 2 months 2021-08-02T19:20:14+00:00 1 Answers 1 views 0

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    2021-08-02T19:21:42+00:00

    Explanation:

    Given that,

    A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical, \theta=15.1^{\circ}

    The magnitude of the magnetic field B changes in time according to the equation :

    B(t)=3.75+2.75 t-7.05 t^2

    Radius of the loop, r = 0.21 m

    We need to find the magnitude of the induced emf in the loop when t=5.63 s. The induced emf is given by :

    \epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA\cos \theta)}{dt}

    B is magnetic field

    A is area of cross section

    \epsilon=A\dfrac{-dB}{dt}\\\\\epsilon=\pi r^2\dfrac{-d(3.75+2.75 t-7.05 t^2)}{dt}\times \cos\theta\\\\\epsilon=\pi r^2\times(2.75-14.1t)\times \cos\theta

    At t = 5.63 seconds,

    \epsilon=-\pi (0.21)^2\times(2.75-14.1(5.63))\times \cos(15.1)\\\\\epsilon=10.25V

    So, the magnitude of induced emf in the loop when t=5.63 s is 10.25 V.

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