A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform mag

Question

A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.

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Nho 3 years 2021-08-22T17:06:49+00:00 1 Answers 1 views 0

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  1. Dau
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    2021-08-22T17:08:28+00:00
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    Answer:

    B = 0.37T

    Explanation:

    In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

    emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}       (1)

    emf: induced voltage in the solenoid = 10,000V

    N: turns of the solenoid = 525

    ФB: magnetic flux

    B: magnitude of the magnetic field = ?

    A: cross-sectional area of the solenoid = π*r^2

    r: radius of the cross-sectional area = 0.260m

    Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

    First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90°  (quarter of a revolution)

    In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

    emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\

    Finally, you replace the values of the parameters to calculate B:

    B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T

    The strength of the magnetic field is 0.37T

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