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: Identify which field component matters.
Loop A lies entirely in the XY plane, so its area vector points along \(\hat z\). Only the \(z\)-component of each magnetic field contributes to the flux linking the loop; the \(\hat x\) and \(\hat y\) components lie in the plane of the loop and give zero flux.

Step 2: Extract the \(z\)-components of the two fields.
From \(\vec{B}_1=\dfrac{m}{\sqrt2}\sin(\omega t)(1\hat x+0\hat y+1\hat z)\), the \(z\)-component is
\[ B_{1z}=\frac{m}{\sqrt2}\sin(\omega t) \]
From \(\vec{B}_2=-\dfrac{n}{\sqrt2}\cos(2\omega t+\pi/4)(0\hat x+1\hat y+1\hat z)\), the \(z\)-component is
\[ B_{2z}=-\frac{n}{\sqrt2}\cos(2\omega t+\pi/4) \]

Step 3: Write the total flux through loop A.
Since \(a_1\) and \(a_2\) are nonintersecting regions inside the same loop, the total flux is the sum of the flux through each region:
\[ \Phi=a_1B_{1z}+a_2B_{2z}=\frac{a_1m}{\sqrt2}\sin(\omega t)-\frac{a_2n}{\sqrt2}\cos(2\omega t+\pi/4) \]

Step 4: Apply Faraday's law to get the induced EMF.
\[ e=-\frac{d\Phi}{dt}=-\frac{a_1m\omega}{\sqrt2}\cos(\omega t)-\frac{2a_2n\omega}{\sqrt2}\sin(2\omega t+\pi/4) \]
This is a sum of two sinusoids, one at frequency \(\omega\) and one at frequency \(2\omega\).

Step 5: Find the rms value of each sinusoid separately.
For a sinusoid of amplitude \(P\), the rms value is \(P/\sqrt2\). The first term has amplitude \(\dfrac{a_1m\omega}{\sqrt2}\), so its rms value is
\[ \frac{1}{\sqrt2}\cdot\frac{a_1m\omega}{\sqrt2}=\frac{a_1m\omega}{2} \]
The second term has amplitude \(\dfrac{2a_2n\omega}{\sqrt2}\), so its rms value is
\[ \frac{1}{\sqrt2}\cdot\frac{2a_2n\omega}{\sqrt2}=a_2n\omega \]

Step 6: Combine the two rms values.
Because the two terms oscillate at different frequencies (\(\omega\) and \(2\omega\)), their product averages to zero over a full cycle, so the mean square values simply add:
\[ e_{rms}^2=\left(\frac{a_1m\omega}{2}\right)^2+(a_2n\omega)^2=\frac{a_1^2m^2\omega^2}{4}+a_2^2n^2\omega^2 \]

Step 7: Combine into a single fraction and take the square root.
\[ e_{rms}^2=\frac{a_1^2m^2\omega^2+4a_2^2n^2\omega^2}{4} \]
\[ \boxed{e_{rms}=\sqrt{\frac{a_1^2\omega^2m^2+4a_2^2\omega^2n^2}{4}}} \]
Hence, the correct option is (A).