Question

An alternating voltage is generated by rotating a coil in a normal magnetic field with 50Hz frequency. If ${V_{rms}} = 220\;{\text{V}}$, then the maximum flux passing through the coil is:
A. $0.99\,{\text{Wb}}$
B. $0.4\,{\text{Wb}}$
C. $0.6\,{\text{Wb}}$
D. $0.7\,{\text{Wb}}$

Answer 1

Use the equation for the root mean square voltage in terms of the peak voltage. Also use the equation for the voltage induced due to changing magnetic flux with time. Integrate the voltage induced with respect to time to obtain the maximum flux.

Formula used:
The formula for the root mean square value of the voltage is
${V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$ …… (1)
Here, ${V_0}$ is the peak value of the voltage.
The induced emf $V$ in the coil is given by
$V = - \dfrac{{d\phi }}{{dt}}$ …… (2)
Here, $d\phi$ is the change in the magnetic flux in time $dt$.
The voltage for the sinusoidal wave is given by
$V = {V_0}\sin 2\pi ft$ …… (3)
Here, ${V_0}$ is the peak voltage and $f$ is the frequency.

Complete step by step answer:
We have given the root mean square voltage $220\;{\text{V}}$.
${V_{rms}} = 220\;{\text{V}}$
The frequency of rotation of the rotating coil is $50\,{\text{Hz}}$.
$f = 50\,{\text{Hz}}$
Rearrange the equation (1) for the peak voltage.
${V_0} = {V_{rms}}\sqrt 2$
Substitute $220\;{\text{V}}$ for ${V_{rms}}$ in the above equation.
${V_0} = \left( {220\;{\text{V}}} \right)\sqrt 2$
$\Rightarrow {V_0} = 220\sqrt 2$
Substitute ${V_0}\sin 2\pi ft$ for $V$ in equation (2).
${V_0}\sin 2\pi ft = - \dfrac{{d\phi }}{{dt}}$
$\Rightarrow \dfrac{{d\phi }}{{dt}} = - {V_0}\sin 2\pi ft$
$\Rightarrow d\phi = - {V_0}\sin 2\pi ftdt$
Integrate both sides of the above equation with respect to time.
$\Rightarrow \int {d\phi } = \int { - {V_0}\sin 2\pi ftdt}$
$\Rightarrow \phi = - {V_0}\int {\sin 2\pi ftdt}$
$\Rightarrow \phi = - {V_0}\left[ { - \dfrac{{\cos 2\pi ft}}{{2\pi f}}} \right]$
$\Rightarrow \phi = {V_0}\dfrac{{\cos 2\pi ft}}{{2\pi f}}$
Substitute $220\sqrt 2$ for ${V_0}$ in the above equation.
$\Rightarrow \phi = \left( {220\sqrt 2 } \right)\dfrac{{\cos 2\pi ft}}{{2\pi f}}$
The flux will be maximum only when the angle $\cos 2\pi ft$ is 1.
Substitute 1 for $\cos 2\pi ft$ in the above equation.
$\Rightarrow \phi = \left( {220\sqrt 2 } \right)\dfrac{1}{{2\pi f}}$
Substitute 3.14 for $\pi$ and $50\,{\text{Hz}}$ for $f$ in the above equation.
$\Rightarrow \phi = \left( {220\sqrt 2 } \right)\dfrac{1}{{2\left( {3.14} \right)\left( {50\,{\text{Hz}}} \right)}}$
$\Rightarrow \phi = 0.99\,{\text{Wb}}$

Therefore, the maximum flux through the coil is $0.99\,{\text{Wb}}$.

Hence, the correct option is A.

Note:
The students should be careful while integrating the equation of voltage for the sinusoidal wave. The students should not forget that after integrating the sine of the angle, they need to also integrate the angle with respect to time and write it in the denominator.