Question

if the resistance of the resistor in the figure is slowly increased, what is the direction of the current induced in the small circular loop inside the larger loop?(figure 1) clockwise counterclockwise part b what would it be if the small loop were placed outside the larger one, to the left? clockwise counterclockwise

Explanation:

Step1: Determine the original magnetic field

According to the right - hand rule, the magnetic field inside the larger loop (due to the current in the larger loop) is directed out of the page.

Step2: Analyze the change in current

When the resistance of the resistor in the larger loop is increased, the current in the larger loop decreases. So, the magnetic flux through the small loop (which is out of the page) decreases.

Step3: Apply Lenz's law (for the small loop inside)

Lenz's law states that the induced current will oppose the change in flux. To oppose the decrease in the out - of - page flux, the induced current in the small loop inside will create a magnetic field out of the page. Using the right - hand rule for a loop, a counter - clockwise current creates a magnetic field out of the page. But wait, this is wrong. Let's re - analyze.

The current in the larger loop is in a direction (say, from the battery, using the right - hand rule for the larger loop, if the current is going around the larger loop in a direction such that the magnetic field inside is out of the page). When the resistance is increased, the current \(I=\frac{\mathcal{E}}{R}\) (where \(\mathcal{E}\) is the emf of the battery and \(R\) is the resistance) decreases. The magnetic flux \(\Phi = B\cdot A\) (where \(B=\mu_0\frac{I}{2r}\) for a circular loop of radius \(r\)) through the small loop (inside) is \(\Phi\propto I\). As \(I\) decreases, to oppose the decrease in \(B\) (out of the page), the induced current in the small loop (inside) will try to increase \(B\). Using the right - hand rule, a clockwise current in the small loop (inside) will create a magnetic field into the page (opposing the decrease of the original out - of - page field in a wrong way. Wait, no. The original field through the small loop (inside) is out of the page. The change is a decrease in out - of - page flux. So the induced current should produce a field in the same direction (out of the page) to oppose the decrease. But for a loop, if we want a field out of the page, the current should be counter - clockwise. But this is a mistake in the initial assumption of the current direction in the larger loop.

Assume the current in the larger loop is such that (using the right - hand rule for the larger loop, with the battery connected, the current is going around the larger loop. Let's assume the current in the larger loop is in a direction that the magnetic field inside the larger loop (where the small loop is) is out of the page. When \(R\) increases, \(I=\frac{\mathcal{E}}{R}\) decreases. The flux \(\Phi = BA\) ( \(B\) is out of the page) through the small loop (inside) decreases. By Lenz's law, the induced current in the small loop (inside) will produce a magnetic field to oppose the decrease. So it will produce a magnetic field out of the page. Using the right - hand rule for the small loop (inside), a counter - clockwise current. But wait, no. Let's use the formula \(\mathcal{E}=-\frac{d\Phi}{dt}\).

The magnetic field due to the larger loop at the center (where the small loop is) is \(B = \frac{\mu_0I}{2R_{large}}\) (assuming the small loop is at the center). The flux through the small loop \(\Phi=B\cdot A_{small}=\frac{\mu_0I A_{small}}{2R_{large}}\). \(\frac{d\Phi}{dt}=\frac{\mu_0A_{small}}{2R_{large}}\frac{dI}{dt}\). Since \(I = \frac{\mathcal{E}}{R}\) and \(\frac{dI}{dt}=-\frac{\mathcal{E}}{R^{2}}\frac{dR}{dt}<0\) (as \(R\) is increasing). The induced emf \(\mathcal{E}=-\frac{d\Phi}{dt}\). Using the right - hand rule for the small loop (inside), the induced current is clockwise.

For the small loop outside (to the left):

Step1: Determine the original magnetic field

The magnetic field due to the larger loop outside (to the left) is into the page (using the right - hand rule for the larger loop's current).

Step2: Analyze the change in current

As \(R\) increases, \(I\) decreases. So the flux through the small loop (outside, to the left) (which is into the page) decreases.

Step3: Apply Lenz's law (for the small loop outside)

By Lenz's law, the induced current will oppose the decrease in the into - the - page flux. So the induced current will create a magnetic field into the page. Using the right - hand rule for a loop, a clockwise current creates a magnetic field into the page.

Answer:

A. clockwise (for the small loop inside)
B. clockwise (for the small loop outside to the left)