A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field \overrightarrow B points into the plane of the paper. At t=0, the loop starts rotating about the common diameter as axis with a constant angular velocity \omega in the magnetic field. Which of the following options is/are correct?

The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude if maximum emf induced in the smaller loop alone, The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper

The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude if maximum emf induced in the smaller loop alone

Electromagnetic Induction PYQ — JEE Advanced 2017 · Shift 1 · Q41 with Solution

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Correct answer: B, D

Key idea: treat the twisted wire as two loops connected in series opposition

Because the wire is twisted into a figure-eight shape, the two loops are traversed in opposite senses when we go around the single closed conductor. Hence the magnetic flux contributions of the two loops have opposite signs.

Let the loop areas be:

smaller loop: $A$
larger loop: $2A$

The magnetic field $\vec B$ is uniform and into the page.

As the whole wire rotates about the common diameter with angular speed $\omega$ , the angle between the area vector of each loop and $\vec B$ changes with time.

If at $t=0$ the plane of the loop is in the plane of the paper, then the area vector is along the normal to the paper, i.e. parallel/antiparallel to $\vec B$ . Therefore for each loop, $\Phi = BA\cos(\omega t)$ up to sign depending on orientation.

Flux through each loop

Choose sign convention so that flux through the smaller loop is positive. Then because of opposite orientation of the larger loop,

$\Phi_1 = BA\cos \omega t$ $\Phi_2 = -B(2A)\cos \omega t$

So total flux linked with the single closed wire is $\Phi_{\text{net}} = \Phi_1 + \Phi_2 = BA\cos\omega t - 2BA\cos\omega t = -BA\cos\omega t$

Induced emf

By Faraday's law, $\mathcal E = -\frac{d\Phi_{\text{net}}}{dt}$

Thus, $\mathcal E = -\frac{d}{dt}(-BA\cos\omega t)$ = -\left(BA\omega\sin\omega t\right)

So the net emf is proportional to $$\sin \omega t$$ with amplitude $$\mathcal E_0 = BA\omega$$ This is exactly the amplitude of emf that would be induced in the smaller loop alone: $$\mathcal E_{\text{small,max}} = BA\omega$$ --- 4. **Check each option** ### Option A > The emf induced in the loop is proportional to the sum of the areas of the two loops If areas simply added, emf amplitude would be proportional to $A+2A=3A$. But here the two loops contribute with opposite signs, so net effective area is $$2A-A = A$$ not $3A$. So **A is false**. --- ### Option B > The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone From above, $$\mathcal E_{\text{net,max}} = BA\omega$$ And for smaller loop alone, $$\mathcal E_{\text{small,max}} = BA\omega$$ Hence they are equal. So **B is true**. --- ### Option C > The net emf induced due to both the loops is proportional to $\cos\omega t$ We found $$\mathcal E \propto \sin\omega t$$ not $\cos\omega t$. So **C is false**. --- ### Option D > The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper Flux is $$\Phi_{\text{net}} \propto \cos\omega t$$ Hence $$\left|\frac{d\Phi}{dt}\right| \propto |\sin\omega t|$$ which is maximum when $$\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$$ At that instant, the area vector is perpendicular to $\vec B$, so the **plane of the loops is parallel to $\vec B$**, i.e. the plane of the loops is **perpendicular to the plane of the paper**. Thus **D is true**. --- 5. **Final answer** Correct options are: $$\boxed{B,\ D}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: B, D

Key idea: treat the twisted wire as two loops connected in series opposition

Let the loop areas be:

smaller loop: $A$
larger loop: $2A$

The magnetic field $\vec B$ is uniform and into the page.

As the whole wire rotates about the common diameter with angular speed $\omega$ , the angle between the area vector of each loop and $\vec B$ changes with time.

Flux through each loop

Choose sign convention so that flux through the smaller loop is positive. Then because of opposite orientation of the larger loop,

$\Phi_1 = BA\cos \omega t$ $\Phi_2 = -B(2A)\cos \omega t$

So total flux linked with the single closed wire is $\Phi_{\text{net}} = \Phi_1 + \Phi_2 = BA\cos\omega t - 2BA\cos\omega t = -BA\cos\omega t$

Induced emf

By Faraday's law, $\mathcal E = -\frac{d\Phi_{\text{net}}}{dt}$

Thus, $\mathcal E = -\frac{d}{dt}(-BA\cos\omega t)$ = -\left(BA\omega\sin\omega t\right)

So the net emf is proportional to $$\sin \omega t$$ with amplitude $$\mathcal E_0 = BA\omega$$ This is exactly the amplitude of emf that would be induced in the smaller loop alone: $$\mathcal E_{\text{small,max}} = BA\omega$$ --- 4. **Check each option** ### Option A > The emf induced in the loop is proportional to the sum of the areas of the two loops If areas simply added, emf amplitude would be proportional to $A+2A=3A$. But here the two loops contribute with opposite signs, so net effective area is $$2A-A = A$$ not $3A$. So **A is false**. --- ### Option B > The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone From above, $$\mathcal E_{\text{net,max}} = BA\omega$$ And for smaller loop alone, $$\mathcal E_{\text{small,max}} = BA\omega$$ Hence they are equal. So **B is true**. --- ### Option C > The net emf induced due to both the loops is proportional to $\cos\omega t$ We found $$\mathcal E \propto \sin\omega t$$ not $\cos\omega t$. So **C is false**. --- ### Option D > The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper Flux is $$\Phi_{\text{net}} \propto \cos\omega t$$ Hence $$\left|\frac{d\Phi}{dt}\right| \propto |\sin\omega t|$$ which is maximum when $$\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$$ At that instant, the area vector is perpendicular to $\vec B$, so the **plane of the loops is parallel to $\vec B$**, i.e. the plane of the loops is **perpendicular to the plane of the paper**. Thus **D is true**. --- 5. **Final answer** Correct options are: $$\boxed{B,\ D}$$ This matches the stored correct answer.

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