Maximum voltage in the primary circuit of a transformer

[lorenz0]

Homework Statement

The primary circuit of a transformer has ##N=240## windings, each one with area ##S=10cm^2##. In the primary circuit flows a sinusoidally alternating current. The magnetic field in the iron core of the transformer goes from the maximum value of ##B_0=65mT## along the direction perpendicular to the windings to the exact same value but in the opposite direction in ##\Delta t=10 ms##.

Find the maximum voltage in the primary circuit.

Relevant Equations

##V=-N\frac{d\phi(\vec{B})}{dt}##, ##V_{max}=\sqrt{2}V_{eff}##

The alternating current will create an induced voltage given by ##V=-N\frac{d\phi(\vec{B})}{dt}=-N\frac{\Delta\phi(\vec{B})}{\Delta t}=N\frac{2 B_0 S}{\Delta t}=\frac{240\cdot 2\cdot 65\cdot 10^{-3}\cdot 10\cdot 10^{-4}}{10\cdot 10^{-3}}=3.12 V## and since this is the effective voltage, the maximum voltage will be ##V_{max}=\sqrt{2}V_{eff}=\sqrt{2}\cdot 3.12 V\approx 4.4 V##. Now, the result given is ##4.9 V## but I haven't been able to figure out where I have made a mistake so I would be grateful if someone could point it out to me, thanks.

 

[BvU]

Homework Statement: The primary circuit of a transformer has ##N=240## windings, each one with area ##S=10cm^2##. In the primary circuit flows a sinusoidally alternating current. ...

You seem to do ##\Delta \Phi/\Delta t## as if it were a sawtooth ...

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[lorenz0]

You seem to do ##\Delta \Phi/\Delta t## as if it were a sawtooth ...

##\ ##

Ah, I think I have understood now. Since ##V(t)=V_0\sin(\omega t)## it must be that ##B(t)=B_0\cos(\omega t)## with ##\omega=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{2\delta t}=100\pi\ \frac{rad}{s}## so that ##V(t)=V_0\sin(\omega t)=-N\frac{d\phi(\vec{B})}{dt}=-N\frac{d}{dt}\left( B_0\cos(\omega t)S )\right)=NB_0S\omega\sin(\omega t)## thus ##V_0=NB_0S\omega=240\cdot 65\cdot 10^{-3}\cdot 10^{-3}\cdot 100\pi\ V\approx 4.9\ V.##