1. A certain elastic conducting material is stretched into a circular loop of 12.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s. What emf is induced in the loop at that instant? (See problem 2, chapter 30 in Fundamentals of Physics)

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**Problem 1: Induced EMF in a Shrinking Circular Loop**

A certain elastic conducting material is stretched into a circular loop of 12.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s. What emf is induced in the loop at that instant? (See problem 2, chapter 30 in Fundamentals of Physics)

*Explanation:*

To find the induced emf (electromotive force) in the loop, we can use Faraday's Law of Induction. According to Faraday's Law, the induced emf is given by:

\[ \mathcal{E} = - \frac{d\Phi_B}{dt} \]

where \( \Phi_B \) is the magnetic flux through the loop. \( \Phi_B \) is given by:

\[ \Phi_B = B \cdot A \]

Here:
- \( B \) is the magnetic field strength (0.800 T).
- \( A \) is the area of the loop. For a circle, \( A = \pi r^2 \), where \( r \) is the radius.

Since the radius \( r \) is changing with time, we need to account for its rate of change (\( \frac{dr}{dt} \)). Therefore:

\[ A = \pi r^2 \]

The rate of change of area with respect to time \( \frac{dA}{dt} \) can be found using the chain rule:

\[ \frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2\pi r \frac{dr}{dt} \]

Given:
- \( r = 12.0 \) cm = 0.120 m
- \( \frac{dr}{dt} = -75.0 \) cm/s = -0.750 m/s (the negative sign indicates that the radius is shrinking)

Then:

\[ \frac{dA}{dt} = 2\pi (0.120 \, \text{m})(-0.750 \, \text{m/s}) \]
\[ \frac{dA}{dt} = -0.180 \pi \, \text{m}^2/\text{s} \]

Finally, the induced emf
Transcribed Image Text:**Problem 1: Induced EMF in a Shrinking Circular Loop** A certain elastic conducting material is stretched into a circular loop of 12.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s. What emf is induced in the loop at that instant? (See problem 2, chapter 30 in Fundamentals of Physics) *Explanation:* To find the induced emf (electromotive force) in the loop, we can use Faraday's Law of Induction. According to Faraday's Law, the induced emf is given by: \[ \mathcal{E} = - \frac{d\Phi_B}{dt} \] where \( \Phi_B \) is the magnetic flux through the loop. \( \Phi_B \) is given by: \[ \Phi_B = B \cdot A \] Here: - \( B \) is the magnetic field strength (0.800 T). - \( A \) is the area of the loop. For a circle, \( A = \pi r^2 \), where \( r \) is the radius. Since the radius \( r \) is changing with time, we need to account for its rate of change (\( \frac{dr}{dt} \)). Therefore: \[ A = \pi r^2 \] The rate of change of area with respect to time \( \frac{dA}{dt} \) can be found using the chain rule: \[ \frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2\pi r \frac{dr}{dt} \] Given: - \( r = 12.0 \) cm = 0.120 m - \( \frac{dr}{dt} = -75.0 \) cm/s = -0.750 m/s (the negative sign indicates that the radius is shrinking) Then: \[ \frac{dA}{dt} = 2\pi (0.120 \, \text{m})(-0.750 \, \text{m/s}) \] \[ \frac{dA}{dt} = -0.180 \pi \, \text{m}^2/\text{s} \] Finally, the induced emf
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