E11. The magnetic flux through a coil of wire changes from $$ 9 \mathrm{~T} \cdot \mathrm{~m}^{2} $$ to zero in a time of 0.50 s . What is the magnitude of the average voltage induced in the coil during this change? E12. A coil of wire with 120 turns and a cross-sectional area of $$ 0.03 \mathrm{~m}^{2} $$ lies with its plane perpendicular to a magnetic field of magnitude 1.8 T . The coil is rapidly removed from the magnetic field in a time of 0.4 s . a. What is the initial magnetic flux through the coil? b. What is the average value of the voltage induced in the coil?

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### **E11. Determining the Average Voltage Induced in a Coil**

**Problem Statement:**
The magnetic flux through a coil of wire changes from $$ 9 \, \text{T} \cdot \text{m}^2 $$ to zero in a time of $$ 0.50 \, \text{s} $$. What is the magnitude of the average voltage induced in the coil during this change?

**Solution:**

To find the induced voltage, we use **Faraday's Law of Electromagnetic Induction**, which states:

$$
\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}
$$

Where:
- $$ \mathcal{E} $$ is the induced electromotive force (emf) in volts (V),
- $$ N $$ is the number of turns in the coil,
- $$ \Delta \Phi $$ is the change in magnetic flux in tesla-square meters (T·m²),
- $$ \Delta t $$ is the time interval in seconds (s).

**Assumptions:**
- The problem mentions "a coil of wire" without specifying the number of turns. Typically, if not specified, assume $$ N = 1 $$.

**Given:**
- Initial magnetic flux, $$ \Phi_{\text{initial}} = 9 \, \text{T} \cdot \text{m}^2 $$
- Final magnetic flux, $$ \Phi_{\text{final}} = 0 \, \text{T} \cdot \text{m}^2 $$
- Time interval, $$ \Delta t = 0.50 \, \text{s} $$

**Calculations:**
1. **Determine the Change in Magnetic Flux:**
   $$
   \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 9 = -9 \, \text{T} \cdot \text{m}^2
   $$
   
2. **Calculate the Induced EMF:**
   $$
   |\mathcal{E}| = | -N \frac{\Delta \Phi}{\Delta t} | = N \left| \frac{\Delta \Phi}{\Delta t} \right|
   $$
   
   Substituting the values:
   $$
   |\mathcal{E}| = 1 \times \left| \frac{-9}{0.50} \right| = \frac{9}{0.50} = 18 \, \text{V}
   $$

**Answer:**
The magnitude of the average voltage induced in the coil is **18 volts**.

---

### **E12. Analyzing a Multi-Turn Coil's Magnetic Flux and Induced Voltage**

**Problem Statement:**
A coil of wire with 120 turns and a cross-sectional area of $$ 0.03 \, \text{m}^2 $$ lies with its plane perpendicular to a magnetic field of magnitude $$ 1.8 \, \text{T} $$. The coil is rapidly removed from the magnetic field in a time of $$ 0.4 \, \text{s} $$.

**Part a. What is the initial magnetic flux through the coil?**

**Solution:**

**Given:**
- Number of turns, $$ N = 120 $$
- Cross-sectional area, $$ A = 0.03 \, \text{m}^2 $$
- Magnetic field, $$ B = 1.8 \, \text{T} $$
- Angle between the magnetic field and the normal to the coil, $$ \theta = 0^\circ $$ (since the plane is perpendicular to $$ \mathbf{B} $$)

**Formula for Magnetic Flux ($$ \Phi $$):**
$$
\Phi = B \cdot A \cdot \cos \theta
$$

Since the magnetic field is perpendicular to the plane of the coil, $$ \cos \theta = \cos 0^\circ = 1 $$.

**Calculations:**
1. **Flux per Turn:**
   $$
   \Phi_{\text{per turn}} = B \cdot A \cdot \cos \theta = 1.8 \times 0.03 \times 1 = 0.054 \, \text{T} \cdot \text{m}^2
   $$
   
2. **Total Magnetic Flux Linkage for the Coil:**
   $$
   \Phi_{\text{total}} = N \times \Phi_{\text{per turn}} = 120 \times 0.054 = 6.48 \, \text{T} \cdot \text{m}^2
   $$

**Answer:**
The initial magnetic flux through the coil is **6.48 T·m²**.

---

**Part b. What is the average value of the voltage induced in the coil?**

**Solution:**

**Given:**
- Time interval for removal, $$ \Delta t = 0.4 \, \text{s} $$
- Initial magnetic flux linkage, $$ \Phi_{\text{initial}} = 6.48 \, \text{T} \cdot \text{m}^2 $$
- Final magnetic flux linkage, $$ \Phi_{\text{final}} = 0 \, \text{T} \cdot \text{m}^2 $$ (since the coil is removed from the field)

**Faraday's Law:**
$$
\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}
$$
For magnitude:
$$
|\mathcal{E}| = N \left| \frac{\Delta \Phi}{\Delta t} \right|
$$

**Calculations:**
1. **Change in Magnetic Flux:**
   $$
   \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - 6.48 = -6.48 \, \text{T} \cdot \text{m}^2
   $$
   
2. **Calculate the Induced EMF:**
   $$
   |\mathcal{E}| = 120 \times \left| \frac{-6.48}{0.4} \right| = 120 \times \frac{6.48}{0.4} = 120 \times 16.2 = 1944 \, \text{V}
   $$
   
   Wait, this result seems excessively high. Let me double-check the calculations.

**Correction:**

Actually, the total flux linkage already accounts for the number of turns. Therefore, the correct formula should be:

$$
|\mathcal{E}| = \left| \frac{\Delta \Phi_{\text{total}}}{\Delta t} \right|
$$

So,

$$
|\mathcal{E}| = \frac{6.48}{0.4} = 16.2 \, \text{V}
$$

**Alternatively:**

If calculating using flux per turn:

$$
|\mathcal{E}| = N \times \left| \frac{\Delta \Phi_{\text{per turn}}}{\Delta t} \right| = 120 \times \frac{0.054}{0.4} = 120 \times 0.135 = 16.2 \, \text{V}
$$

**Answer:**
The average induced voltage in the coil is **16.2 volts**.

---

**Summary of Answers:**

- **E11:** The induced voltage is **18 V**.
- **E12a:** The initial magnetic flux is **6.48 T·m²**.
- **E12b:** The average induced voltage is **16.2 V**.
- **E11:** The induced voltage is **18 volts**. - **E12a:** The initial magnetic flux is **6.48 T·m²**. - **E12b:** The average induced voltage is **16.2 V**.

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