A rectangular coil with sides 0.130 m by 0.250 m has 513 turns of wire. It is rotated about its long axis in a uniform magnetic field of 0.550 T that is perpendicular to the long axis. At what frequency, in complete cycles per second, must the coil be rotated for it to generate a maximum voltage of 115 V?

voltage = N * Δ(BA)/Δt
BA = 0.57*0.16*0.22 = 2.0064e-2
N = 505
115/505 = Δ(BA)/Δt = 23/101
When the top of the coil rotates to the bottom (1/2 half cycle) BA changes from max to min and when the bottom rotates back to the top BA changes from min to max. So Δ(BA) is twice per cycle
So 2*101Δ(BA)=23Δt and Δt = 1/f
202*2.0064e-2/23 =Δt = 1/f => f =5.675Hz

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shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-04-13T08:39:13+02:00

The frequency in complete cycles per second, must the coil be rotated for it to generate a maximum voltage of 115 V will be f=5.675 Hertz

What is frequency?

Frequency is defined as the number of cycles achieved in per second.It is generally an inverse of time period.

Now the voltage is given as

[tex]voltage = \dfrac{N \times \Delta (BA)}{\Delta t}[/tex]

[tex]BA = 0.57\times0.16\times0.22 = 2.0064e^{-2}[/tex]

N = 505

When the top of the coil rotates to the bottom (1/2 half cycle) BA changes from max to min and when the bottom rotates back to the top BA changes from min to max. So Δ(BA) is twice per cycle

So[tex]2\times 101\Delta(BA)=23\Delta t[/tex]

[tex]202\times 2.0064e^{\dfrac{-2}{23}} =\Delta t = \dfrac{1}{f} = > f =5.675Hz[/tex]

Hence the frequency in complete cycles per second, must the coil be rotated for it to generate a maximum voltage of 115 V will be f=5.675 Hertz

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