An MRI technician moves her hand from a region of very low magnetic field strength ( approximately zero) into the MRI scanner's field in a quick motion, taking 3.1 milliseconds. The MRI is essentially a solenoid, with 1500 loops of wire. By doing so, a current of 0.15 milliAmps is induced in her wedding ring. If her wedding ring has a diameter of 1.95 cm and it has measured resistance of 119 Ohms, what is the magnetic field in the MRI.

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An MRI technician moves her hand from a region of very low magnetic field strength ( approximately zero) into the MRI scanner's field in a quick motion, taking 3.1 milliseconds. The MRI is essentially a solenoid, with 1500 loops of wire. By doing so, a current of 0.15 milliAmps is induced in her wedding ring. If her wedding ring has a diameter of 1.95 cm and it has measured resistance of 119 Ohms, what is the magnetic field in the MRI. 

**Title: Understanding Magnetic Fields in MRI Technology**

**Problem Explanation:**

An MRI technician moves her hand from an area with low magnetic field strength into the field of an MRI scanner. This movement occurs over 3.1 milliseconds. The scenario involves a 0.15-milliamp current induced in a wedding ring, which has a 1.95 cm diameter, being wound with 1500 loops of wire. The ring’s resistance is 119 Ohms. The goal is to determine the magnetic field in the MRI.

**Illustrations:**

- A simple diagram shows a hand moving into an MRI scanner, indicating the directional axes (x, y, z).
- The magnetic field is depicted with lines, and a coil (representing the wedding ring) is shown with loops around a central point.

**Variables and Equations:**

- \(\Delta t = 3.1 \, \text{ms}\)
- \(I = 0.15 \, \text{mA}\)
- \(d = 1.95 \, \text{cm}\)
- \(R = 119 \, \Omega\)
- \(N = 1500\)
- \(B = ?\)

Relevant Equations:
- Ohm's Law: \(V = IR\)
- Electromagnetic Force: \(\mathcal{E} = \frac{\Delta \Phi}{\Delta t}\)
- Magnetic Flux: \(\Phi_i = BA \cos \Theta\)

**Solution Steps:**

1. **Voltage Calculation:**
   \[
   V = IR = (0.15 \times 10^{-3})(119) = 0.01785 \, \text{V} = \mathcal{E}
   \]

2. **Magnetic Flux:**
   \[
   \Phi_i = BA \cos \Theta = B (\pi (1.95)^2) = 11.9459B
   \]

3. **Total Flux:**
   \[
   \Phi_{\text{tot}} = N \Phi_i = (1500) (11.9459B) = 17918.859B
   \]

4. **Electromotive Force:**
   \[
   \mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{17918.859B}{3.1 \times
Transcribed Image Text:**Title: Understanding Magnetic Fields in MRI Technology** **Problem Explanation:** An MRI technician moves her hand from an area with low magnetic field strength into the field of an MRI scanner. This movement occurs over 3.1 milliseconds. The scenario involves a 0.15-milliamp current induced in a wedding ring, which has a 1.95 cm diameter, being wound with 1500 loops of wire. The ring’s resistance is 119 Ohms. The goal is to determine the magnetic field in the MRI. **Illustrations:** - A simple diagram shows a hand moving into an MRI scanner, indicating the directional axes (x, y, z). - The magnetic field is depicted with lines, and a coil (representing the wedding ring) is shown with loops around a central point. **Variables and Equations:** - \(\Delta t = 3.1 \, \text{ms}\) - \(I = 0.15 \, \text{mA}\) - \(d = 1.95 \, \text{cm}\) - \(R = 119 \, \Omega\) - \(N = 1500\) - \(B = ?\) Relevant Equations: - Ohm's Law: \(V = IR\) - Electromagnetic Force: \(\mathcal{E} = \frac{\Delta \Phi}{\Delta t}\) - Magnetic Flux: \(\Phi_i = BA \cos \Theta\) **Solution Steps:** 1. **Voltage Calculation:** \[ V = IR = (0.15 \times 10^{-3})(119) = 0.01785 \, \text{V} = \mathcal{E} \] 2. **Magnetic Flux:** \[ \Phi_i = BA \cos \Theta = B (\pi (1.95)^2) = 11.9459B \] 3. **Total Flux:** \[ \Phi_{\text{tot}} = N \Phi_i = (1500) (11.9459B) = 17918.859B \] 4. **Electromotive Force:** \[ \mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{17918.859B}{3.1 \times
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