The image presents a problem involving a parallel plate capacitor moving through a magnetic field. The problem statement is as follows:---The drawing shows a parallel plate capacitor that is moving with a speed of 41.2 m/s through a 3.40-Tesla magnetic field. The velocity \( v \) is perpendicular to the magnetic field. The electric field within the capacitor has a value of 139 N/C, and each plate has an area of \( 7.91 \times 10^{-4} \, \text{m}^2 \). What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?---**Diagram Description:**- The diagram shows a parallel plate capacitor with positive and negative charges on opposite plates.- Several horizontal arrows representing the electric field lines are directed from the positive plate to the negative plate.- The magnetic field is indicated by red arrows pointing perpendicularly to the motion direction.**Calculation Notes:**The input field shows a calculated result:- Number: \( 2.353 \times 10^{-10} \)- Units: Newtons (N)**Additional Information:**The tolerance for the answer is \(\pm 1\) in the 3rd significant digit.

The charge on the positive plate of the capacitor is .The magnitude of the magnetic force exerted on…

The drawing shows a parallel plate capacitor that is moving with a speed of 41.2 m/s through a 3.40-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 139 N/C, and each plate has an area of 7.91 × 10-4 m². What is the magnitude of the magnetic force exerted on the positive plate of the capacitor? The tolerance is ± 1 in the 3rd significant digit. Number 2.353E-10 Units N

The drawing shows a parallel plate capacitor that is moving with a speed of 41.2 m/s through a 3.40-T magnetic field. The velocity v is perpendicular to the magnetic field. The electric field within the capacitor has a value of 139 N/C, and each plate has an area of 7.91 × 10-4 m². What is the magnitude of the magnetic force exerted on the positive plate of the capacitor? The tolerance is ± 1 in the 3rd significant digit. Number 2.353E-10 Units N

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

Question

The image presents a problem involving a parallel plate capacitor moving through a magnetic field. The problem statement is as follows:

---

The drawing shows a parallel plate capacitor that is moving with a speed of 41.2 m/s through a 3.40-Tesla magnetic field. The velocity \( v \) is perpendicular to the magnetic field. The electric field within the capacitor has a value of 139 N/C, and each plate has an area of \( 7.91 \times 10^{-4} \, \text{m}^2 \). What is the magnitude of the magnetic force exerted on the positive plate of the capacitor?

---

**Diagram Description:**

- The diagram shows a parallel plate capacitor with positive and negative charges on opposite plates.
- Several horizontal arrows representing the electric field lines are directed from the positive plate to the negative plate.
- The magnetic field is indicated by red arrows pointing perpendicularly to the motion direction.

**Calculation Notes:**

The input field shows a calculated result:
- Number: \( 2.353 \times 10^{-10} \)
- Units: Newtons (N)

**Additional Information:**

The tolerance for the answer is \(\pm 1\) in the 3rd significant digit.

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