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 ...

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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 $V_0=NB_0S\omega=240\cdot 65\cdot 10^{-3}\cdot 10^{-3}\cdot 100\pi\ V\approx 4.9\ V.$