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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?

Verified Answer :

voltage = N Δ(BA)/Δt 
BA = 0.570.160.22 = 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 2101Δ(BA)=23Δt and Δt = 1/f 
202*2.0064e-2/23 =Δt = 1/f => f =5.675Hz

shubhamchouhanvrVT

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

\(voltage = \dfrac{N \times \Delta (BA)}{\Delta t}\)

\(BA = 0.57\times0.16\times0.22 = 2.0064e^{-2}\)

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\(2\times 101\Delta(BA)=23\Delta t\)

\(202\times 2.0064e^{\dfrac{-2}{23}} =\Delta t = \dfrac{1}{f} = > f =5.675Hz\)

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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