Question

The figure shows an arbitrarily shaped planar conducting loop A in the XY plane. Two nonintersecting regions with areas \(a_1\) and \(a_2\) within the loop are subjected to magnetic fields \(\vec{B}_1=\dfrac{m}{\sqrt2}\sin(\omega t)\left(1\,\hat{x}+0\,\hat{y}+1\,\hat{z}\right)\), and \(\vec{B}_2=-\dfrac{n}{\sqrt2}\cos(2\omega t+\pi/4)\left(0\,\hat{x}+1\,\hat{y}+1\,\hat{z}\right)\), respectively.

What is the expression for the induced rms voltage in loop A?

• A. \(\sqrt{\dfrac{a_1^2\omega^2m^2+4a_2^2\omega^2n^2}{4}}\)
• B. \(\sqrt{\dfrac{a_1^2\omega^2m^2+4a_2^2\omega^2n^2}{2}}\)
• C. \(\sqrt{\dfrac{a_1^2\omega^2m^2-2a_2^2\omega^2n^2}{2}}\)
• D. \(\sqrt{a_1^2\omega^2m^2+2a_2^2\omega^2n^2}\)

Hint:

Only the z-component of each field links flux through this planar loop; combine the two frequencies using mean-square addition, not simple addition of amplitudes.

Answer 1

The Correct Option is A

Step 1: Reduce the fields to the useful component.
The loop sits flat in the $xy$ plane, so its normal is along $\hat z$ and only the $z$ parts of $\vec B_1$ and $\vec B_2$ produce flux. These are $B_{1z}=\dfrac{m}{\sqrt2}\sin(\omega t)$ and $B_{2z}=-\dfrac{n}{\sqrt2}\cos(2\omega t+\pi/4)$.

Step 2: Build the flux and the EMF.
\[ \Phi(t)=a_1\cdot\frac{m}{\sqrt2}\sin(\omega t)-a_2\cdot\frac{n}{\sqrt2}\cos(2\omega t+\pi/4) \]
\[ e(t)=-\frac{d\Phi}{dt}=-\frac{a_1m\omega}{\sqrt2}\cos(\omega t)-\sqrt2\,a_2n\omega\sin(2\omega t+\pi/4) \]
using $\dfrac{2}{\sqrt2}=\sqrt2$.

Step 3: Square the EMF and average over one period directly.
Let $A=\dfrac{a_1m\omega}{\sqrt2}$ and $B=\sqrt2\,a_2n\omega$, so $e(t)=-A\cos(\omega t)-B\sin(2\omega t+\pi/4)$. Then
\[ e(t)^2=A^2\cos^2(\omega t)+B^2\sin^2(2\omega t+\pi/4)+2AB\cos(\omega t)\sin(2\omega t+\pi/4) \]

Step 4: Average each term over a full period.
The time average of $\cos^2(\omega t)$ is $\dfrac12$, and the time average of $\sin^2(2\omega t+\pi/4)$ is also $\dfrac12$. The cross term is a product of sinusoids at different frequencies ($\omega$ and $2\omega$); expanding it as a sum of sinusoids at frequencies $\omega$ and $3\omega$ shows every piece averages to zero over a full period.

Step 5: Write the mean square value.
\[ \langle e^2\rangle=\frac{A^2}{2}+\frac{B^2}{2}=\frac12\left(\frac{a_1^2m^2\omega^2}{2}\right)+\frac12\left(2a_2^2n^2\omega^2\right)=\frac{a_1^2m^2\omega^2}{4}+a_2^2n^2\omega^2 \]

Step 6: Take the square root to get the rms value.
\[ e_{rms}=\sqrt{\frac{a_1^2m^2\omega^2}{4}+a_2^2n^2\omega^2}=\sqrt{\frac{a_1^2\omega^2m^2+4a_2^2\omega^2n^2}{4}} \]
This matches option (A).
\[ \boxed{e_{rms}=\sqrt{\dfrac{a_1^2\omega^2m^2+4a_2^2\omega^2n^2}{4}}} \]