A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 22.0 cm long and has a diameter of 2.50 cm. At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1600 A/s. ▼ Part A For this time, calculate the average magnetic flux through each turn of the inner solenoid. Express your answer in webers. 15 ΑΣΦ Submit Part B Request Answer M = ? For this time, calculate the mutual inductance of the two solenoids. Express your answer in henries. 17 ΑΣΦ Wb ? H

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Chapter1: Units, Trigonometry. And Vectors
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**Part C**

For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.  
Express your answer in volts.

Box with icons and a text input field:  
\[ \mathcal{E}_2 = \]  [Text input box] V

Buttons:  
- Submit
- Request Answer

This section prompts the user to calculate the electromagnetic force (emf) in volts induced in an outer solenoid due to a changing current in an inner solenoid. The text input box allows the user to enter their answer, and there are options to submit the response or request the correct answer.
Transcribed Image Text:**Part C** For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid. Express your answer in volts. Box with icons and a text input field: \[ \mathcal{E}_2 = \] [Text input box] V Buttons: - Submit - Request Answer This section prompts the user to calculate the electromagnetic force (emf) in volts induced in an outer solenoid due to a changing current in an inner solenoid. The text input box allows the user to enter their answer, and there are options to submit the response or request the correct answer.
**A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 22.0 cm long and has a diameter of 2.50 cm. At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1600 A/s.**

---

### Part A

For this time, calculate the average magnetic flux through each turn of the inner solenoid.

**Express your answer in webers.**

\[ \Phi_B = \ \_\_\_\_ \, \text{Wb} \]

- Input box for answer
- Options: Submit and Request Answer

---

### Part B

For this time, calculate the mutual inductance of the two solenoids.

**Express your answer in henries.**

\[ M = \ \_\_\_\_ \, \text{H} \]

- Input box for answer
- Options: Submit and Request Answer
Transcribed Image Text:**A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 22.0 cm long and has a diameter of 2.50 cm. At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1600 A/s.** --- ### Part A For this time, calculate the average magnetic flux through each turn of the inner solenoid. **Express your answer in webers.** \[ \Phi_B = \ \_\_\_\_ \, \text{Wb} \] - Input box for answer - Options: Submit and Request Answer --- ### Part B For this time, calculate the mutual inductance of the two solenoids. **Express your answer in henries.** \[ M = \ \_\_\_\_ \, \text{H} \] - Input box for answer - Options: Submit and Request Answer
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