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This question is motivated by Problem 7.12 in Griffiths Electrodynamics book. I have not included it in the homework section, because I have already solved it correctly. However, I question whether my solution which agrees with the solution's manual is correct.

Relevant Equations:

$$\Phi = \int{\vec{B}\cdot d\vec{a}}$$

$$\mathcal{E} = -\frac{d\Phi}{dt}$$

$$I = \mathcal{E}/R$$

The problem states: A long solenoid, of radus a, is driven by an alternating current, so that the field inside is $$\vec{B(t)}{= B_0cos(wt)}\hat{z}$$ A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

My solution was to take the field of the solenoid, integrate it over the loop, take the time derivative and use the relation $$\mathcal{E} = -\frac{d\Phi}{dt}$$ Then, one simply takes $$I = \frac{\mathcal{E}}{R}$$ This works just fine, and you'll get the correct answer: $$\frac{\pi a^2wB_0sin(wt)}{4R}$$

However, my concern is that we are ignoring the self inductance. Because the current in the loop is changing, this also creates an emf. The equation we should start with is $$\Phi = \int{\vec{B}\cdot d\vec{a}}$$ taken over the loop's surface. But, the field is no longer the field due simply to the solenoid, but the field should include the contribution from the loop itself because it now has some current running that is changing in time. This would be enormously more complicated as it's hard to calculate the field off axis of a loop. But, shouldn't we include this in the flux calculation?

Relevant Equations:

$$\Phi = \int{\vec{B}\cdot d\vec{a}}$$

$$\mathcal{E} = -\frac{d\Phi}{dt}$$

$$I = \mathcal{E}/R$$

The problem states: A long solenoid, of radus a, is driven by an alternating current, so that the field inside is $$\vec{B(t)}{= B_0cos(wt)}\hat{z}$$ A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

My solution was to take the field of the solenoid, integrate it over the loop, take the time derivative and use the relation $$\mathcal{E} = -\frac{d\Phi}{dt}$$ Then, one simply takes $$I = \frac{\mathcal{E}}{R}$$ This works just fine, and you'll get the correct answer: $$\frac{\pi a^2wB_0sin(wt)}{4R}$$

However, my concern is that we are ignoring the self inductance. Because the current in the loop is changing, this also creates an emf. The equation we should start with is $$\Phi = \int{\vec{B}\cdot d\vec{a}}$$ taken over the loop's surface. But, the field is no longer the field due simply to the solenoid, but the field should include the contribution from the loop itself because it now has some current running that is changing in time. This would be enormously more complicated as it's hard to calculate the field off axis of a loop. But, shouldn't we include this in the flux calculation?

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