**Finding the Magnitude of \(\oint \vec{B} \cdot d \vec{s}\) for the Closed Path (a)**In the problem presented, we need to find the magnitude of \(\oint \vec{B} \cdot d \vec{s}\) for the closed path labeled (a) shown in the diagram.**Diagram Explanation:**The diagram depicts a field with three currents flowing through it: \(1A\) (1 Ampere), \(5A\) (5 Amperes), and \(2A\) (2 Amperes). These are marked by circular crosses which typically indicate currents going into the plane of the diagram. The path (a) is represented by a dashed, closed loop.**Problem Statement:**Based on the given diagram, we are asked to compute the magnitude of \(\oint \vec{B} \cdot d \vec{s}\), where \(\vec{B}\) is the magnetic field and \(d \vec{s}\) represents an infinitesimal vector element of the path (a).**Options Provided:**a. \(4 \mu_0 \) (T.m) b. \(3 \mu_0 \) (T.m) c. \(1 \mu_0 \) (T.m) d. \(0\) To solve this, we utilize Ampère's Law which states that the integral of the magnetic field \(\vec{B}\) around a closed path is equal to \(\mu_0\) times the total current enclosed by that path:\[ \oint \vec{B} \cdot d \vec{s} = \mu_0 I_{\text{enc}} \]Here, \( I_{\text{enc}} \) is the net current enclosed by path (a). Let's determine \( I_{\text{enc}} \):1. Current \(5A\) is going into the plane (positive direction as per the assumption).2. Current \(2A\) is also going into the plane (positive direction).3. Current \(1A\) is going into the plane (positive direction).Thus,\[ I_{\text{enc}} = 5A + 2A + 1A = 8A \]Using Ampère's Law:\[ \oint \vec{B} \cdot d \vec{s}

Answered: Image /qna-images/answer/1acffed3-4afb-492e-a3d9-1db412f69788.jpg

Answered: Find the magnitude of 08 .d3 for the…

Find the magnitude of 08 .d3 for the closed path (a )shown in the figure: 05 A 02 A Select one: a. 4H. (T.m) O b. 3. (Tm) C. Ip. (Tm) d. o

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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Question

**Finding the Magnitude of \(\oint \vec{B} \cdot d \vec{s}\) for the Closed Path (a)**

In the problem presented, we need to find the magnitude of \(\oint \vec{B} \cdot d \vec{s}\) for the closed path labeled (a) shown in the diagram.

**Diagram Explanation:**

The diagram depicts a field with three currents flowing through it: \(1A\) (1 Ampere), \(5A\) (5 Amperes), and \(2A\) (2 Amperes). These are marked by circular crosses which typically indicate currents going into the plane of the diagram. The path (a) is represented by a dashed, closed loop.

**Problem Statement:**

Based on the given diagram, we are asked to compute the magnitude of \(\oint \vec{B} \cdot d \vec{s}\), where \(\vec{B}\) is the magnetic field and \(d \vec{s}\) represents an infinitesimal vector element of the path (a).

**Options Provided:**

a. \(4 \mu_0 \) (T.m)
b. \(3 \mu_0 \) (T.m)
c. \(1 \mu_0 \) (T.m)
d. \(0\)

To solve this, we utilize Ampère's Law which states that the integral of the magnetic field \(\vec{B}\) around a closed path is equal to \(\mu_0\) times the total current enclosed by that path:

\[ \oint \vec{B} \cdot d \vec{s} = \mu_0 I_{\text{enc}} \]

Here, \( I_{\text{enc}} \) is the net current enclosed by path (a).

Let's determine \( I_{\text{enc}} \):

1. Current \(5A\) is going into the plane (positive direction as per the assumption).
2. Current \(2A\) is also going into the plane (positive direction).
3. Current \(1A\) is going into the plane (positive direction).

Thus,

\[ I_{\text{enc}} = 5A + 2A + 1A = 8A \]

Using Ampère's Law:

\[ \oint \vec{B} \cdot d \vec{s}

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