**Problem 5:**- **Objective:** Find the size and direction of the force per meter exerted on the bottom wire by the top 4 wires. Current \( I = 5 \, \text{A} \) in all wires. The distance between each wire is \( 1 \, \text{cm} \).- **Task:** Also find the strength of the magnetic field at the location of the middle wire.- **Additional Requirement:** Find the magnetic field (\( \mathbf{B} \)) at the location of the 2nd wire from the top due to the other 4 wires.**Formulae:**- Force per unit length (\( F/l \)) between two parallel currents: \[ F/l = \frac{{\mu_0 \times I_1 \times I_2}}{{2\pi \times d}} \]- Magnetic field (\( B \)) due to a long straight current-carrying wire: \[ B = \frac{{\mu_0 \times I}}{{2\pi \times r}} \]**Diagram:**- **Description:** The diagram consists of five parallel horizontal lines, each representing a wire. All arrows point to the right, indicating the direction of the current. The wires are equally spaced.**Solution Space:**- **Force per Meter (\( F/l \)) Calculation:** \[ F/l = \underline{\phantom{F/l =}} \, \text{N/m} \hspace{0.5cm} (up, down, left, or right) \]- **Magnetic Field (\( B \)) Calculation at the 2nd Wire:** \[ B = \underline{\phantom{B =}} \, \text{T} \hspace{0.5cm} (up, down, left, or right) \]- **Magnetic Field (\( B \)) at the Middle Wire:** \[ B = \underline{\phantom{B =}} \, \text{T} \hspace{0.5cm} (up, down, left, or right) \]

Answered: Image /qna-images/answer/9c35bf25-6d05-4b31-995e-a25ef9f01841.jpg

5. 9. Find the size and direction of the force per meter exerted on the bottom wire by the top 4. I-5A, in all wires. The distance between each wire is lcm. Also find the strength of the magnetic seld at the location of the middie wire. FIND THE 4dLo, r, 1 AT LOCATI0N Of 2 WIRG PROA TOp pue TOTHE F/A= N/m UTHER 4- T. l ת LdL ,o B=

5. 9. Find the size and direction of the force per meter exerted on the bottom wire by the top 4. I-5A, in all wires. The distance between each wire is lcm. Also find the strength of the magnetic seld at the location of the middie wire. FIND THE 4dLo, r, 1 AT LOCATI0N Of 2 WIRG PROA TOp pue TOTHE F/A= N/m UTHER 4- T. l ת LdL ,o B=

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 Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

**Problem 5:**

- **Objective:** Find the size and direction of the force per meter exerted on the bottom wire by the top 4 wires. Current \( I = 5 \, \text{A} \) in all wires. The distance between each wire is \( 1 \, \text{cm} \).

- **Task:** Also find the strength of the magnetic field at the location of the middle wire.

- **Additional Requirement:** Find the magnetic field (\( \mathbf{B} \)) at the location of the 2nd wire from the top due to the other 4 wires.

**Formulae:**

- Force per unit length (\( F/l \)) between two parallel currents:

\[
F/l = \frac{{\mu_0 \times I_1 \times I_2}}{{2\pi \times d}}
\]

- Magnetic field (\( B \)) due to a long straight current-carrying wire:

\[
B = \frac{{\mu_0 \times I}}{{2\pi \times r}}
\]

**Diagram:**

- **Description:** The diagram consists of five parallel horizontal lines, each representing a wire. All arrows point to the right, indicating the direction of the current. The wires are equally spaced.

**Solution Space:**

- **Force per Meter (\( F/l \)) Calculation:**

\[
F/l = \underline{\phantom{F/l =}} \, \text{N/m} \hspace{0.5cm} (up, down, left, or right)
\]

- **Magnetic Field (\( B \)) Calculation at the 2nd Wire:**

\[
B = \underline{\phantom{B =}} \, \text{T} \hspace{0.5cm} (up, down, left, or right)
\]

- **Magnetic Field (\( B \)) at the Middle Wire:**

\[
B = \underline{\phantom{B =}} \, \text{T} \hspace{0.5cm} (up, down, left, or right)
\]

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